numerical connections —...

CHAPTER

15Numerical connections — stoichiometry

You will examine:

• mass–mass stoichiometry

• measurement of concentration of solutions and ions in solutions (in mol L–1)

• mass–concentration–volume stoichiometry reactions in solution

• pH calculations

• titrations

• quantitatively interpret balanced chemical equations

• solve mass–mass stoichiometry problems

• calculate molarity

• solve stoichiometric problems involving solutions

• calculate quantities of reactants and products based on balanced chemical equations

What features would you look for when

you buy your first vehicle? Although

colour, design and performance

may rate highly on your list, many

people place safety features as

their first priority. In new car models,

for example, airbags have become

almost mandatory. Airbags rely on a

chemical reaction to inflate the bags.

How do chemists know how much of

the reactant chemicals to place in the

airbag for correct inflation?

UNIT 2 Environmental chemistry336

Calculations using balanced equations

How is chemistry involved in the operation of an automotive airbag? Before being inflated, airbags contain a small amount of solid sodium azide, NaN3. On impact with another vehicle or object, an electric spark is released, acting as a trigger. Very quickly, the sodium azide in the airbag decomposes to form solid sodium, Na, and nitrogen gas, N2, according to the chemical equation:

2NaN3(s) 2Na(s) + 3N2(g)

The sodium then reacts with water vapour in air. The nitrogen gas produced in the reaction inflates the airbag in time to absorb much of the impact energy of the crash, which may have caused injury to the driver. The sodium azide reaction must produce just the right amount of nitrogen gas to inflate the bag to the correct pressure. How do designers know how much sodium azide to use so that the airbag is correctly inflated? They use stoichiometric calculations.

The study of relative amounts — ratios — of substances involved in chemical reactions is known as stoichiometry. The word ‘stoichiometry’ comes from two Greek words meaning ‘element’ and ‘measure’. Knowledge of stoichiometry is essential whenever quantitative information about a chemical reaction is required.

Stoichiometry is used every day in the home and in industry. Baking a cake in the kitchen requires a recipe to ensure proper quantities of all ingredients are used. Imagine how a cake would taste if a chef added three times the required amount of flour! Many cooks have disasters in the kitchen by not using ingredients in the correct ratios. On a larger scale, an industrial chemist reducing iron in a blast furnace needs to know exactly how much iron ore, Fe2O3, coke, C, and limestone (calcium carbonate), CaCO3, to feed into the top of the furnace in order to produce a certain amount of iron. A disaster in industry can be life-threatening because incorrect ratios of chemicals can result in uncontrolled, violent explosions.

Solving stoichiometric problems is particularly important in fields such as food chemistry, drug chemistry, forensic science and in any industry where chemicals are manufactured. Stoichiometry may be thought of as the means of obtaining a recipe for a chemical reaction.

What a balanced chemical equation tells us

Chemically speaking, if we wish to say, ‘One molecule of nitrogen gas reacts with three molecules of hydrogen gas to form two molecules of ammonia gas,’ we simply write:

N2(g) + 3H2(g) 2NH3(g)

It would be impossible and impractical to make ammonia molecule by molecule. Chemists work with larger quantities called moles (symbol mol). The coefficients in a balanced chemical equation tell us the relative numbers of moles of reactants and products taking part in the reaction.

1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas.

The equation actually indicates the mole ratio of reactants and products. The beauty of this method is that the ratios never change. If 1 mole of nitrogen reacts with 3 moles of hydrogen, then 2 moles of nitrogen will react with 6 moles of hydrogen, 0.5 mole of nitrogen will react with 1.5 moles of hydrogen, and so on.

Assuming that there is plenty of nitrogen, how much ammonia will be produced by 3 moles of hydrogen? The equation tells us: 2 moles of ammonia. Similarly, 6 moles of hydrogen will produce 4 moles of ammonia.

Stoichiometry is the term used for calculations based on chemical equations. Stoichiometry predicts the relative amounts of substances involved in chemical reactions.

Stoichiometry is the term used for calculations based on chemical equations. Stoichiometry predicts the relative amounts of substances involved in chemical reactions.

Chemical equations show the relative number of molecules participating in a chemical reaction. They can also show the relative number of moles, the mole ratio, of each substance involved.

Chemical equations show the relative number of molecules participating in a chemical reaction. They can also show the relative number of moles, the mole ratio, of each substance involved.

337CHAPTER 15 Numerical connections — stoichiometry

An equation also tells us about the ratio between the reactants and the products of other quantities. It tells us about the ratio between the number of particles, and the masses of the reactants and products.

So, when 1 mole of N2 gas reacts with 3 moles of H2 gas to produce 2 moles of NH3 gas, we know that this can be restated as: 28 g of N2 gas (or 6.02 s 1023 N2 molecules) reacts with 6 g of H2 gas (or 1.806 s 1024 H2 molecules) to produce 34 g of NH3 gas (or 1.204 s 1024 NH3 molecules).

What a chemical equation does not tell us

An equation conveys no information about the rate of a reaction. It may be fast, as in the explosive oxidation of hydrogen to form water; or it may be slow, as in the oxidation or rusting of iron.

An equation does not tell us whether a reaction requires heat or gives off heat, or what temperature or pressure is needed.

Finally, an equation gives no details as to how the individual atoms or molecules are transformed from reactants to products. This information is essential for full understanding of how a chemical reaction takes place.

Sample problem: 15.1 Using an equation to calculate moles of reactants used and products

formed

Methanol, CH3OH, is used as a fuel in some racing cars. The equation for the combustion of methanol in the car’s engine is:

2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(g)

(a) How many moles of oxygen are needed to react with 4 moles of methanol?

(b) How many moles of water are formed when 1 mole of oxygen reacts completely with methanol?

Solution:

(a) The equation tells us that 2 moles of methanol will react with exactly 3 moles of oxygen. Use this ratio (2 : 3) to calculate that 4 moles of methanol will react with 6 moles of oxygen exactly.

(b) The equation tells us that 3 moles of oxygen will produce 4 moles of water. Use the ratio (3 : 4) to conclude that 1 mole of oxygen will produce 1 1

3 moles

of water.

++

The balanced chemical equation for

the formation of ammonia can be

interpreted in several ways.

The balanced chemical equation for

the formation of ammonia can be

interpreted in several ways.

A chemical equation does not indicate the rate (how fast) or extent (how far) of a reaction.

A chemical equation does not indicate the rate (how fast) or extent (how far) of a reaction.

Different fuels are suited for

different purposes. High-

performance racing cars may run

on diesel, ethanol, methanol or

petrol. All of these fuels react in

oxygen to produce carbon dioxide

and water vapour.

Different fuels are suited for

different purposes. High-

performance racing cars may run

on diesel, ethanol, methanol or

petrol. All of these fuels react in

oxygen to produce carbon dioxide

and water vapour.


Revision questions

1. Silver tarnishes partly because of the presence of small amounts of hydrogen sulfi de, H2S (a gas that originates from the decay of food and smells like rotten eggs), according to the reaction:

4Ag(s) + 2H2S(g) + O2(g) 2Ag2S(s) + 2H2O(l)

(a) How many moles of silver sulfi de form from the complete reaction of 1 mole of silver?

(b) How many moles of hydrogen sulfi de will react with 1 mole of silver?(c) How many moles of silver sulfi de form from 3.5 moles of hydrogen

sulfi de?

2. Methane is the main component of natural gas that we use as a fuel. When methane burns in air, the following reaction takes place:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

(a) How many moles of methane are needed to react with 1 mole of oxygen gas?

(b) How many moles of oxygen are needed to react with 0.1 moles of methane?

(c) How many moles of carbon dioxide are produced from 0.1 moles of methane?

(d) How many moles of water are produced from 0.1 moles of methane?(e) How many moles of carbon dioxide will be produced by 0.1 moles of

oxygen gas?(f) How many moles of oxygen gas will react completely with 0.25 moles of

methane?(g) How many moles of water will be produced from 8 moles of oxygen gas?

Mass – mass stoichiometry

Mass–mass stoichiometry involves solving a problem in which the mass of a reactant or product is given. You are then asked to calculate the mass of another reactant or product. Mass–mass stoichiometry requires the con-version of masses of substances to moles, or moles of substances to masses. Such conversions make use of the formulae:

molesmass

molar mass

mass mo

( )( )

( )

( )

nm

M

m

�

� lles molar mass( ) ( )n Ms

Mass–mass stoichiometric problems can be solved in four steps:

1. Write a balanced chemical equation for the reaction, identifying the known (given) and unknown (required) quantities of substance.

2. Calculate the number of moles of the known quantity of substance present.

3. From the equation, fi nd the molar ratio that states the proportion of known to unknown quantities in the reaction and use it to calculate the number of moles of the required substance.

4. Calculate the quantity (mass) of the required substance.

Mass–mass stoichiometry involves calculating the mass of one of the substances in a reaction by using the equation and the mass of one of the other substances. The steps are:• Write a balanced equation,

identifying the known and unknown quantities.

• Change equation coeffi cients into moles.

• From the equation, use the molar ratio to predict the moles of the required substance.

• Change the moles into mass of the required substance.

Mass–mass stoichiometry involves calculating the mass of one of the substances in a reaction by using the equation and the mass of one of the other substances. The steps are:• Write a balanced equation,

identifying the known and unknown quantities.

• Change equation coeffi cients into moles.

• From the equation, use the molar ratio to predict the moles of the required substance.

• Change the moles into mass of the required substance.

‘Fool’s gold’ is chemically

known as pyrites, FeS2.

‘Fool’s gold’ is chemically

known as pyrites, FeS2.


Sample problem: 15.2 Calculating the mass of a substance given the mass of another

reactant or product

Some sulfur is present in coal in the form of pyrites, FeS2. This substance is also known as ‘fool’s gold’. When pyrites burns, it pollutes the air with one of its combustion products, sulfur dioxide, and produces solid iron(III) oxide. What mass of iron(III) oxide is formed from the complete combustion of 183.5 g of pyrites?

Solution:

Step 1 Write the equation, identifying the known and unknown quantities of substance.

4FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)

Pyrites is the known quantity and iron(III) oxide is the unknown quantity.

Step 2 Calculate the number of moles of the known quantity of substance, FeS2.

n( ).

. ( . )

..

.

FeS2183 5

55 85 2 32 1

183 5120 051 5

�

� s

�

� 229 mol

Step 3 Find the molar ratio, FeS2:Fe2O3, from the equation, and use it to calculate the number of moles of the required substance, Fe2O3.

The ratio is 4 : 2, which becomes 2 : 1.

n nn

( ) ( )( )

.

.

Fe O FeSFeS

2 3 222

4 21 529

20 7643

� s �

�

� mmol

Step 4 Calculate the required quantity (mass) of Fe2O3 formed by the reaction.

m n M( )

. ( . )

.

Fe O

122 1 g

r2 3

0 7643 2 55 85 3 16

� s

� s s � s

�

Revision questions

3. Calculate the mass of water that is produced when 2.8 g of methane is burnt in air.

4. The reaction that occurs in the reusable solid rocket booster for modern space shuttles is shown in the following equation:

3Al(s) + 3NH4ClO4(s) Al2O3(s) + AlCl3(s) + 3NO(g) + 6H2O(g)

Calculate the mass of each product formed in a reaction when 100 g of aluminium is completely used up.


5. Phosphine, PH3, is a poisonous gas once used as pest control for stored grain and may be prepared according to the following equation:

Ca3P2(s) + 6H2O(l) 3Ca(OH)2(s) + 2PH3(g)

Calculate the mass of phosphine that can be prepared from 25 g of Ca3P2.

6. Propane burns in air according to the equation:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

(a) What mass of water will be produced by 6.5 g of oxygen?

(b) What mass of O2 will react with 1.7 g of C3H8?

(c) What mass of CO2 will be produced by 0.50 moles of propane gas, C3H8?

(d) How many grams of propane gas will be needed to produce 5.92 g of CO2(g)?

(e) How many kilograms of CO2(g) are released into the atmosphere when the entire contents of a 5.0 kg cylinder of propane are used at a barbecue?

7. Industrially, metallic iron may be obtained from an ore such as haematite, Fe2O3, after reaction in a blast furnace according to the equation:

Fe2O3(s) + 3CO(g) 2Fe(l) + 3CO2(g)

Use a spreadsheet program or a graphics calculator to set up a spreadsheet to calculate the mass of iron that may be formed by this reaction, given an initial mass of haematite. The following template can be used as the basis for your spreadsheet.

A B C D

1 mass of haematite used (g)

moles of haematite formed

moles of iron formed

mass of iron formed (g)

2

• In cell B2, type = A2/159.7, then click and drag from bottom right corner of cell down the column. Fill down.

• In cell C2, type = 2*B2, then click and drag from bottom right corner of cell down the column.

• In cell D2, type = C2*55.85, then click and drag from bottom right corner of cell down the column.

Once your spreadsheet has been set up, you can enter values for the mass of haematite used in the reaction, and let your spreadsheet calculate the amount of iron you would expect to be produced.

(a) What mass of iron would be produced by 23.7 kg of haematite?

(b) Explain the formulas set up in cells B2, C2 and D2.

(c) Add columns to your spreadsheet to calculate the mass of carbon dioxide that may be formed in the reaction.

(d) What mass of carbon dioxide would be produced by 805 kg of haematite?


Limiting reactant calculations

When the amounts of reactants provided for a reaction are not in the mole ratio, amounts of one or more reactants will be left over once the reaction has taken place.

The amounts of the products will be limited by the amount of the reactant that is completely used up in the reaction. To return to our previous example:

N2(g) + 3H2(g) 2NH3(g)

Recall that a balanced equation indicates the mole ratios in which the reactants are used up and the products are formed. According to this equation one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas.

If one mole of nitrogen and four moles of hydrogen are mixed and allowed to react, then all the nitrogen and three moles of hydrogen are used up. This means that one mole of hydrogen is left over. The reactant that is completely used up (in this case nitrogen) is called the limiting reactant. Any unreacted reactants (hydrogen in this example) are called excess reactants. The amount of product formed by the reaction is limited by the amount of the limiting reactant. In this case the addition of more hydrogen to the mixture will have no effect on the amount of product formed, as there is no more nitrogen available to react with it.

Sample problem: 15.3 Determining the limiting reactant to calculate the mass of a product

Magnesium and oxygen react to form magnesium oxide according to the fol-lowing equation:

2Mg(s) + O2(g) 2MgO(s)

If 24.31 g of magnesium and 20.00 g of oxygen are available for reaction, determine the limiting reactant; the reactant in excess and by how many mole and the mass of magnesium oxide formed.

Solution:

Step 1 Convert the given information to moles.

n

n

( )..

.

( )..

.

Mg 1 00 mol

O 0

� �

� �

24 3124 3120 0032 02 6625 mol

Step 2 Compare this mole ratio with the mole ratio in the equation. According to the equation, 2 mol of magnesium reacts with 1 mol of oxygen. Therefore, if 1.00 mol of Mg were to react we would need 0.500 mol of O2. We have 0.625 mol O2, which is more than enough. Therefore, Mg is the limiting reactant.

Step 3 The amount of product formed is determined by the amount of limit-ing reactant. According to the equation, 2 mol of magnesium produces 2 mol of magnesium oxide. Therefore, 1.00 mol mg would produce 1.00 mol MgO.

Step 4 Calculate the mass of MgO produced.

m = 1.00 × M(MgO)

= 1.00 × 40.3

= 40.3 g.

In a limiting reagent calculation, first determine which substance is completely used up (the limiting reagent) and which is left over (the excess reagent). Calculations are then based on the moles of the limiting reagent.

In a limiting reagent calculation, first determine which substance is completely used up (the limiting reagent) and which is left over (the excess reagent). Calculations are then based on the moles of the limiting reagent.


This can be shown in a table:

Equation 2Mg(s) + O2(g) 2MgO(s)

Mole ratio 2 1 2

Initial mass 24.31 g 20.00 g ?

Initial no. of mol 1.00 mol 0.625 mol ?

No. of mol needed 1.00 mol 0.500 mol Produces 1.00 mol

No. of mol in excess 0.625 – 0.500 = 0.125 mol

Mass 24.31 g 0.500 × 32 = 16 g 40.3 g

From the initial 20 g of O2, 16 g is used and 4 g is in excess.

Revision questions

8. Elemental fluorine is very difficult to prepare by ordinary chemical reactions. It was not until 1986 that the following chemical reaction producing fluorine was reported:

2K2MnF6(aq) + 4SbF5(s) 4KSbF6(aq) + 2MnF3(s) + F2(g)

If 5.32 g of K2MnF6 is mixed with 8.75 g of SbF5, calculate the mass of elemental fluorine produced.

9. Mercury reacts with bromine to produce mercury(II) bromide according to the equation:

Hg(l) + Br2(l) HgBr2(s)

If 2.15 g of Hg is present with 1.56 g of Br2, calculate:

(a) the mass of HgBr2 produced(b) the mass of the reactant that remains in excess.

Concentrating on solutions

Many chemical reactions take place between substances that have been dissolved in a liquid, most commonly water, to form a solution. The liquid in which the substances are dissolved is called the solvent; the dissolved substance is called the solute.

When working with solutions, the most easily measured quantity is their volume. However, for the volume of a solution to provide useful information, the solution’s concentration — that is, the amount of solute in a given vol-ume — must be known.

Calculating concentrations of solutions

Concentration can be expressed in a number of ways. Chemists most com-monly use molar concentration or molarity, which is the amount of solute, in moles, present in each litre of solution. The symbol for concentration is c. The unit of measurement for molar concentration is moles per litre, which can be expressed as mol L–1 or sometimes M. Therefore a 1.5 M solution has a concen-tration of 1.5 mol L–1.

15.1 Preparation of a solution of known concentration15.1 Preparation of a solution of known concentration

The concentration of a solution is the number of moles that it contains in each litre. The concentration of a solution is calculated using the formula c n

V� .

This can be transposed to find the number of moles, n = cV.

The concentration of a solution is the number of moles that it contains in each litre. The concentration of a solution is calculated using the formula c n

V� .

This can be transposed to find the number of moles, n = cV.


If 1 mol of solute is dissolved in a total volume of 1 L of water, the concen-tration of the solution is 1 mol L–1. If 0.5 mol of solute is dissolved in a total volume of 1 L of water, the concentration of the solution is 0.5 mol L–1. If, however, 0.5 mol of solute is dissolved in a total volume of 0.5 L of water, the concentration of the solution is 1 mol L–1.

Molar concentration is defined as:

concentrationquantity of solute moles

( )(

c �))

( )

(

volume of solution L

molesvolume

i

c

cnV

�

� nn mol L 1 )

The formula for concentration can be manipulated to calculate the amount, and hence the mass, of the solute.

n = c s V (Note: Volume must be in litres.)

m = n × molar mass

In order to prepare a particular volume of solution of known concentration, the following five steps should be followed:

1. Calculate the number of moles of solute that are needed to obtain the correct concentration of solution for the volume of solvent to be used, according to the formula n = cV.

2. Calculate the mass of the solute needed, using the formula m = nM.

3. Partially fill a volumetric flask with water, and add the correct mass of solute.

4. Dissolve the solute.

5. Add water to the required volume.

Sample problem: 15.4 Preparing a solution of known concentration and volume

Calculate the number of moles of sodium chloride needed to prepare 500 mL of a 0.0800 mol L–1 salt solution. What mass of sodium chloride would be weighed out?

Solution:

Step 1 List the known information.

volume (V) = 0.50 L

concentration (c) = 0.0800 mol L–1

Step 2 Calculate the number of moles (n) of NaCl needed.

n c V� s

� s

�

0 0800 0 500. .

.0 0400 mol

Step 3 Calculate the mass represented by the number of moles.

m n� s

� s

�

molar mass

2 34 g

0 0400 58 5. .

.

The preparation of a solution of

known concentration. To make a

0.5 mol L–1 solution, add 0.5 mole

of solute to a 1.0 litre volumetric

flask half filled with water. Swirl the

flask carefully to make the solute

dissolve. Finally, fill the flask with

water exactly to the 1.0 L mark.

The preparation of a solution of

known concentration. To make a

0.5 mol L–1 solution, add 0.5 mole

of solute to a 1.0 litre volumetric

flask half filled with water. Swirl the

flask carefully to make the solute

dissolve. Finally, fill the flask with

water exactly to the 1.0 L mark.

1 L

0.5 mol solute

1 L

0.5 mol solute


Revision questions

10. Calculate the concentration (molarity) of copper(II) sulfate, CuSO4, in:(a) 1 L of solution containing 200 g CuSO4·5H2O(b) 2.0 L of solution containing 250 g CuSO4·5H4O(c) 700 mL of solution containing 750 g CuSO4·5H2O(d) 125 mL of solution containing 50 g CuSO4·5H2O.

The concentration of water

All pure liquids have a concentration, since concentration can be defined as the number of particles in a given volume. Given that it is possible to mix two liquids together and get a smaller volume, it follows that empty space exists between the liquid molecules. Although to the naked eye water appears to have no gaps, at the molecular level there is always empty space (vacuum) between the water molecules. So liquid water has a concentration.

The concentration of water can be expressed as grams per litre (density) or moles per litre.

Given that the density of water is 1 g mL–1, it follows that 1000 mL of water has a mass of 1000 g. This can be converted to moles using the formula

nmM� . The molar mass of water (M) is 18 g mol–1. Therefore

n( )

.

H O

55 5 mol

21000

18�

�

One litre of pure water therefore contains 55.5 mol of water molecules.

Calculating the concentration of ions in solution

Many ionic substances are soluble. The process of dissolving involves the ionic lattice breaking up and the anions and cations dissociating from each other.

The concentrations of the resulting ions can be calculated. These concen-trations are called ionic concentrations and they are designated by square brackets. For example:

[Na+] = 0.5 mol L–1 means that the ionic concentration of Na+ ions in solution is 0.5 mol L–1.

Sample Problem: 15.5 Calculating the concentration of ions in a solution

Sodium carbonate, Na2CO3, is often used as an ingredient in washing powders because it softens hard water. It dissolves readily in water: 10.6 g of sodium carbonate is dissolved in 500 mL of water.

(a) What is the concentration of the solution?(b) What is the concentration of each ion in the solution?

Solution:

(a)

Step 1 Calculate the number of moles.

nmM

( ).

.

Na CO2 310 61060 100

� �

�

Sodium bicarbonate tablet

dissolving in water, releasing

carbon dioxide bubbles. Many

chemical reactions occur in water.

Sodium bicarbonate tablet

dissolving in water, releasing

carbon dioxide bubbles. Many

chemical reactions occur in water.

Na2CO

3(s)

Na2CO

3(s)

H2O

2Na+

(aq) + CO3

2–(aq)

Na+

Na +

Na+

Na+

CO3

2–

CO3

2–

Na2CO

3(s)

Na2CO

3(s)

H2O

2Na+

(aq) + CO3

2–(aq)

Na+

Na +

Na+

Na+

CO3

2–

CO3

2–

Dissolving sodium carbonate brings

about dissociation of the sodium

and carbonate ions.

Dissolving sodium carbonate brings

about dissociation of the sodium

and carbonate ions.

Ionic concentrations are calculated by finding the concentration of the solution, then multiplying the mole ratio of ions in the dissociation equation.

Ionic concentrations are calculated by finding the concentration of the solution, then multiplying the mole ratio of ions in the dissociation equation.


Step 2 Calculate the concentration.

cnV

( ).

.

.

Na CO

mol L

2 3

1

0 10 500

0 200

� �

�

(b)

Step 1 Write the dissociation equation: H2O

Na2CO3(s) 2Na+(aq) + CO32–(aq)

Step 2 Write the mole ratios.

1 mol Na2CO3 produces 2 mol Na+ and 1 mol CO32–

Step 3 Use mole ratios to determine ion concentration as the volume is constant.

c(Na+) = 2 s c(Na2CO3) = 2 s 0.200 = 0.400 mol L–1

c(CO32–) = c(Na2CO3) = 0.200 s 1 = 0.200 mol L–1

Sample problem: 15.6 Calculating the concentration of a solution given the concentration

of an ion in solution

In an Al2(SO4)3 solution the concentration of SO42– is 0.050 mol L–1. What is the

concentration of solute?

Solution:

Step 1 Write the dissociation equation.H2O

Al2(SO4)3(s) 2Al3+(aq) + 3SO42–(aq)

Step 2 Compare the mole ratios. Since 1 mol Al2(SO4)3 produces 3 mols of SO4

2–

n

n

n n

( ( ) )

(SO )

(Al (SO ) ) (SO

2 4 3

4

2 4 3

Al SO2

13

13

�

� s 44 )2

Step 3 Since the volume is constant,

c c( ( ) ) ( )

.

Al SO SO

0 050 mol L

0

42 4 32

1

1313

� s

� s

�

..

.

0166 mol L

0 017 mol L

�

1

1

Revision questions

11. Caustic soda, NaOH, is an ingredient in household drain cleaners. If a 250 mL bottle of cleaner contains 20 g of caustic soda, calculate:(a) the concentration of the solution(b) the concentration of each ion in the solution.

12. Calculate the concentration of an Mg(NO3)2 solution in which the concentration of nitrate ions is 0.030 mol L–1.

13. Calculate the concentration of a solution of ammonium phosphate, (NH4)3PO4 in which the concentration of ammonium ions is 1.25 mol L–1.


Dilution

When a solution is diluted by the addition of more solvent (e.g. water), the number of moles of solute remains the same. The addition of water to a con-centrated solution does not alter the number of moles or the mass of the solute in that solution.

After a solution has been diluted, the number of moles of solute remains the same. So, if n1 represents the number of moles of the initial or concentrated solution and n2 represents the number of moles of the final or dilute solution, then we can say that:

n1 = n2

Using the equation n = c s V, we can write equations for n1 and n2 as follows:

n1 = c1 s V1 and n2 = c2 s V2

Since the values of n1 and n2 are equal, these equations can be combined to form the following formula:

c1 s V1 = c2 s V2

Sample problem: 15.7 Using the dilution formula to calculate the volume changes

What volume of 10 mol L–1 stock hydrochloric acid is needed to prepare 250 mL of a 2.0 mol L–1 hydrochloric acid solution?

Solution:

Step 1 Let c1 and V1 represent the quantities of concentrated solution and c2 and V2 represent the quantities of dilute solution.

V1 = ? L V2 = 0.250 L

c1 = 10. 0 mol L–1 c2 = 2.0 mol L–1

Step 2 Calculate the quantity of 10 mol L–1 HCl needed to produce the required dilution.

c V c V

V

V

1 1 2 2

1

1

10 0 2 0 0 250

2 0 0 25010 0

�

s � s

�s

�

. . .

. ..

00 050 L

50 mL

.

�

Revision questions

14. Calculate the volume of stock solution needed to prepare 250 mL of a 2.15 mol L–1 solution of hydrochloric acid (hydrochloric acid stock solution = 12 mol L–1).

15. Calculate the concentration (molarity) of 2.0 L of 2.0 mol L–1 solution after dilution with 500 mL of water.

Dilution of a solution will change its concentration. As the volume increases with dilution, the concentration decreases: c1V1 = c2V2.

Dilution of a solution will change its concentration. As the volume increases with dilution, the concentration decreases: c1V1 = c2V2.


Step 3

Stopper the flask and mix thoroughly, then add more H2O to the 100 mL mark. The original 0.200 mol L–1 K2Cr2O7 has been diluted by a factor of 5, so the con-centration decreases by a factor of 5. You now have 100 mL of 0.0400 mol L–1

K2Cr2O7.

Mass–concentration stoichiometry

Many chemical reactions involve interactions between solids and solutions. These reactions may be referred to as mass–concentration stoichiometry. Two solutions may be mixed to form a precipitate or solids may dissolve in some solutions to form new products. Stoichiometric calculations that involve solids and solutions require the use of two formulae to calculate moles. When dealing with solids, we use the formula:

nmM�

When dealing with solutions, we use the formula:

n = cV

Mass–concentration stoichiometric problems may be solved in four steps:1. Write a balanced chemical equation for the reaction, identifying the known

(given) and unknown (required) quantities of substance.2. Calculate the number of moles of the known quantity of substance

present.3. From the equation, find the molar ratio that states the proportion of known

to unknown quantities in the reaction and use it to calculate the number of moles of the required substance.

4. Calculate the quantity of the required substance.Note that these steps are essentially the same as for solving mass–mass

stoichiometry problems.

Mass – volume stoichiometry involves the same steps as mass–mass stoichiometry, except that the formula n = cV is used.

Mass – volume stoichiometry involves the same steps as mass–mass stoichiometry, except that the formula n = cV is used.

Step 1

Measure 20 mL of 0.20 mol L–1 K2Cr2O7 into a pipette. Slowly add the K2Cr2O7 to a calibrated flask.

Step 2

Rinse the pipette with water and also pour it into the calibrated 100 mL flask.


Sample problem: 15.8 Using stoichiometry to calculate quantities

Hydrofluoric acid etches glass and quartz according to the reaction:

SiO2(s) + 4HF(aq) SiF4(g) + 2H2O(l)

A quartz sculptor has a 500 mL container of 22.50 mol L–1 hydrofluoric acid. What mass of quartz could be etched by the acid?

Solution:


SiO2 (s) + 4HF(aq) SiF4 + 2H2O(1) unknown known

Step 2 Calculate the number of moles of the known quantity of substance.

n cV( )

. .

.

HF

11 3 mol

�

� s

�

22 50 0 500

Step 3 Find the molar ratio from the equation and use it to calculate the number of moles of SiO2 required.

n(SiO2) : n(HF) is 1 : 4

n n( ) ( )

.

SiO HF

2 81 mol

214�

�

Step 4 Calculate the required quantity of SiO2, using m = n s M.

m� s � s

�

2 81 28 1 2 16. ( . )

163 g

Sample problem: 15.9 Using experimental data to calculate quantities

A piece of aluminium is placed in a beaker containing 500 mL of H2SO4 solu-tion and hydrogen gas is evolved. Given that the initial mass of Al was 15.14 g and its final mass was 9.74 g, calculate the concentration of the acid.

Solution:


2Al (s) + 3H2SO4(aq) Al2(SO4)3(aq) + 6H2(g) known unknown

Step 2 Determine the mass of Al that was used in the reaction and calculate the number of moles this mass represents.

mass Al

5 40 g

Al

0 20

( ) . .

.

( ).

.

�

�

�

�

15 14 9 74

5 4027

n

mol

Hydrofluoric acid may be used

to decorate crystal glasses and

decorative items or etch markings

into laboratory glassware. This

acid can be kept in plastic reagent

bottles.

Hydrofluoric acid may be used

to decorate crystal glasses and

decorative items or etch markings

into laboratory glassware. This

acid can be kept in plastic reagent

bottles.


Step 3 Find the molar ratio from the equation and use it to calculate the number of moles of H2SO4 required.

n(H2SO4) : n(Al) = 3 : 2

n(H2SO4) n( )

.

.

Al

0 30 mol

3232

0 20

�

� s

�

Step 4 Calculate the concentration of the H2SO4 required for the reaction.

cnV

( )

.

.

.

H SO

0 600 mol L

2 4

1

0 3000 500

�

�

�

Revision questions

16. Zinc metal is reacted with 400 mL of a 0.250 mol L–1 solution of sulfuric acid, H2SO4. Calculate the mass of zinc sulfate formed.

17. Consider the following balanced equation:

Cd(NO3)2(aq) + Na2S(aq) CdS(s) + 2NaNO3(aq)

(a) Calculate the mass of CdS produced from 235 mL of a 0.178 mol L–1 solution of Na2S with excess Cd(NO3)2 present.

(b) What is the significance of Cd(NO3)2 being present in excess?

18. In a laboratory experiment, a strip of magnesium weighing 2.56 g was placed into 200 mL of a hydrochloric acid solution. Bubbles of gas were observed and identified as hydrogen. The next day, no bubbles were observed in the beaker so the magnesium strip was removed from the beaker, dried and reweighed. Its mass was recorded as 0.350 g.(a) Why were no bubbles observed in the beaker on the second day of the

experiment?(b) Calculate the concentration of the acid.(c) Identify the sources of error in this experiment.(d) Outline the safety precautions that should be taken in this experiment.

Solution stoichiometryWhen some solutions are mixed, chemical reactions may occur. Solu-tion stoichiometry, sometimes also called ‘concentration–concentration’ stoichiometry, involves reactions in solution such as precipitation reactions and the neutralisation of an acid and a base to form water and a salt. Solution stoichiometry utilises the formula:

n = cV

to calculate moles of solute in solutions which may react or which may be produced. Solution stoichiometry problems can be solved in four steps:1. Write a balanced chemical equation for the reaction, identifying the known

(given) and unknown (required) quantities of substance.2. Calculate the number of moles of the known quantity of substance present,

using the formula n = cV.

Solution stoichiometry also uses the same steps as for mass–mass stoichiometry. The difference is that the formula n = cV is used for both the given and the required substances.

Solution stoichiometry also uses the same steps as for mass–mass stoichiometry. The difference is that the formula n = cV is used for both the given and the required substances.


3. From the equation, find the molar ratio of known to unknown quantities in the reaction and use it to calculate the number of moles of the required substance.

4. Calculate the quantity (i.e. concentration) of the required substance.

Sample problem: 15.10 Calculating volumes and concentrations of reacting solutions

A solution of cloudy ammonia is analysed for its ammonium hydroxide concentration. A 20.00 mL sample is neutralised with 30.00 mL of 1.10 mol L–1 hydrochloric acid. Find the molarity or concentration of the ammonium hydroxide in the cloudy ammonia.

Solution:


NH4OH(aq) + HCl(aq) NH4Cl(aq) + H2O(l)

c1 = ? mol L–1 c2 = 1.10 mol L–1

V1 = 0.02000 L V2 = 0.03000 L

Step 2 Calculate the number of moles of the known quantity of HCl.

n c V( )

. .

.

HCl

0 0330 mol

� s

� s

�

2 2

1 10 0 03000

Step 3 Find the molar ratio from the equation and use it to calculate the number of moles of the required substance.

n (NH4OH) : n (HCl) = 1 : 1

n (NH4OH) = n (HCl)

Step 4 Calculate the quantity (concentration) of the required substance.

cnV

( ).

.

.

NH OH

1 65 mol L

4

1

0 03300 02000

� �

�

Revision questions

19. In an experiment to determine the concentration of a solution of hydrochloric acid, 15.5 mL of the HCl was neutralised by 25.0 mL of a 0.055 mol L–1 solution of Na2CO3 in a conical flask.(a) Write a balanced equation for the reaction.(b) Determine the concentration of the hydrochloric acid.

20. What volume of a solution of 0.200 mol L–1 potassium hydroxide, KOH, will exactly neutralise 20.0 mL of a solution of 0.300 mol L–1 hydrochloric acid?

21. What volume of a solution of 0.200 mol L–1 potassium hydroxide, KOH, will exactly neutralise 20.0 mL of a solution of 0.300 mol L–1 sulfuric acid, H2SO4?

22. What volume of a solution of 0.200 mol L–1 potassium hydroxide, KOH, will exactly neutralise 20.0 mL of a solution of 0.300 mol L–1 phosphoric acid, H3PO4?

Clouds of condensed ammonia hide

the surface of Saturn. On Earth,

ammonia exists in gaseous form.

Ammonia gas dissolved in water

is called ammonia it is a common

ingredient in many kitchen and

bathroom cleaning agents.

Clouds of condensed ammonia hide

the surface of Saturn. On Earth,

ammonia exists in gaseous form.

Ammonia gas dissolved in water

is called ammonia it is a common

ingredient in many kitchen and

bathroom cleaning agents.


23. The salinity of seawater can be measured by adding silver nitrate solution until all the chloride ions have been precipitated as silver chloride according to the equation:

Ag+(aq) + Cl–(aq) AgCl(s)

A 20.00 mL sample of seawater required 22.40 mL of 0.500 mol L–1 silver nitrate solution to precipitate all the chloride ions from solution. Calculate the molarity of the sodium chloride in the sample.

Acid–base titrations

Acid–base titrations are based on neutralisation reactions. They are a type of volumetric analysis where the unknown concentration of a solution is determined by reacting it with a solution of known concentration. The solution of known concentration is called a standard solution and is prepared by dissolving an accurate amount of solute in water using a volumetric flask that is calibrated to contain the specified volume. The standard solution is placed in a burette, which is used to deliver definite but variable volumes of liquid. In a titration, the volume of liquid measured by the burette is called a titre. The solution of unknown concentration is added to a conical flask using a pipette. A pipette is used to deliver a known volume of liquid which is then called an aliquot.

(d)

burette

and stand

0 mL

50 mL

(c)

20 mL

pipette

(b)

conical

flask

(a)

250 mL

volumetric flask

Apparatus used in titration

A suitable indicator is added to the aliquot in the conical flask. The solution in the burette is slowly added to the aliquot until the indicator changes colour. This process is called titration. The point at which chemically equivalent amounts of acid and base (according to the equation) are present is called the equivalence point. The point at which the indicator changes colour is called the end point and is usually about one drop after the equivalence point. Titres within 0.05 mL of each other are called concordant titres. Three concordant titres are needed to calculate the average titre.

A standard solution of hydrochloric acid is difficult to prepare because the concentrated form of hydrochloric acid produces fumes and there are dangers involved in trying to measure out a very accurate volume. For this reason hydrochloric acid is usually standardised with a known concentration of a suitable base. This involves a titration and allows accurate concentrations of hydrochloric acid to be determined. A standard solution of sodium carbonate is sometimes used as the base as its preparation is relatively easy.

Acid–base titration is an experimental method that uses specialised equipment to determine the concentration of a solution. A requirement is a solution of known concentration with which to react the solution of unknown concentration. Solutions whose concentrations are known accurately are called standard solutions.

Acid–base titration is an experimental method that uses specialised equipment to determine the concentration of a solution. A requirement is a solution of known concentration with which to react the solution of unknown concentration. Solutions whose concentrations are known accurately are called standard solutions.

The burette should be read at eye

level to obtain an accurate reading.

The burette should be read at eye

level to obtain an accurate reading.

To use an interactive simulation

of acid–base titration, go to the

website for this book and click

on the Acid–base simulation

weblink (see Weblinks,

page 532).

To use an interactive simulation

of acid–base titration, go to the

website for this book and click

on the Acid–base simulation

weblink (see Weblinks,

page 532).


Sample problem: 15.11 Calculating the concentration of a standard solution

1.33 g of anhydrous sodium carbonate is placed in a 250.0 mL (0.25 L) volu-metric flask and water is added to the required mark.

This standard solution is placed in the burette and titrated against a solution of hydrochloric acid whose accurate concentration is to be determined. It was found that a 20.00 mL aliquot of HCl required a Na2CO3 titre of 23.50 mL for complete neutralisation.

Calculate the concentration of:(a) the standard solution of Na2CO3

(b) the HCl solution.

Solution:

(a) Use cnV� to calculate the concentration of the standard solution.

n

cnV

( ).

.

( )

.

Na CO

Na CO

2 3

2 3

1 331060 0155

0 0155

�

�

�

�

00 2500

0 0620 1

.

.� mol L

(b) Write the equation for the reaction, use it to find the molar ratio, then calculate the concentration of the solution.

Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

n c V

n

( )

. .

( ) .

Na CO

HCl

2 3

0 0620 0 02350

2 0 0620

� s

� s

� s ss

� s

� �s

0 02350

2 91 10

2 91 100

3

3

.

.

( ).

.

mol

HClcnV 002000

4 1�

0 1 6 mol L.

Therefore the hydrochloric acid has a molarity of 0.146 mol L–1.

Revision questions

24. (a) Sodium oxalate, Na2C2O4, can be used to make a standard solution. Describe how this is done.

(b) If 3.183 g of sodium oxalate is dissolved in water in a 250 mL volumetric flask, find the concentration of the solution.

25. (a) 1.461 g of dry sodium carbonate, Na2CO3, is dissolved in water in a 250 mL volumetric flask. (i) Why can this be called a standard solution?(ii) Calculate the concentration of the solution.

(b) 20 mL aliquots of this solution were titrated with nitric acid, HNO3, three times. The average concordant titre was found to be 22.17 mL. (i) Write the equation for the reaction of sodium carbonate with nitric

acid.(ii) Find the concentration of the nitric acid.


Calculating pH values

The pH scale can be used to compare the acidity or basicity of different solutions. As acidity of aqueous solutions is due to the presence of H3O+, the pH is calculated from the concentration of H3O+. Mathematically, pH can be found by taking the logarithm to base 10 of the hydrogen (or hydronium) ion concentration and then adding a negative sign.

pH = –log10[H3O+] or [H3O+] = 10–pH

For example, to find the pH of an aqueous solution in which [H3O+] is 0.01 mol L–1 first express [H3O+] as a power of 10 before calculating the logarithm, then multiply by –1, as shown below:

[ ] .

[ ]

H O mol L

mol L

log H O

31

2 1

10 3

0 01

10

�

�

�

�

� llog

pH

10210

2

1 2

2

[ ]

( )

�

� s

�

This means that when the [H3O+] in a solution is expressed as a power of 10 the pH is equal to the power multiplied by –1.

If [H3O+] = 10–x, then pH = x.

Therefore

[H3O+] = 10–pH

Sample problem: 15.12 Calculating the pH of (a) monoprotic and (b) diprotic acidic solutions

(a) A 0.01 mol L–1 solution of HNO3 is prepared. Calculate the pH of the solution.

(b) A 0.005 mol L–1 solution of H2SO4 is prepared. Assuming complete ionisation, calculate the pH of the solution.

Solution:

(a) Since HNO3 is a strong acid, it will ionise completely according to the equation:

HNO3(aq) + H2O(l) H3O+(aq) + NO3–(aq)

Therefore:

c c( ) ( ) .

[ ]

HNO H O 0 01 mol L

H O mol

3 31

3210

� �

�

�

� LL

pH

< �

1

2

(b) Since H2SO4 is a strong acid, it will ionise according to the equation:

H2SO4(aq) + 2H2O(l) 2H3O+(aq) + SO4–(aq)

If 1 mole of H2SO4 produces 2 moles of H3O+. So:

c c( ) ( )

.

.

[

H O H SO

0 01 mol L

H O

3 2 4

1

3

2

2 0 005

�

� s

� s

�

��

< �

] 10

2

2 1mol L

pH

The pH of a solution may be calculated by using the formula:

pH = –log10[H3O+]

The pH of a solution may be calculated by using the formula:

pH = –log10[H3O+]

This Galileo probe image of the

atmosphere of Venus shows sulfuric

acid clouds.

This Galileo probe image of the

atmosphere of Venus shows sulfuric

acid clouds.


Sample Problem: 15.13 Calculating the pH of a dilution

50 mL of a solution of HCl of pH 2 is diluted so that the new pH is 5. What volume of water was added?

Solution:

Step 1 Convert the pH measurements to molarity.

pH 2 [H3O+] = 10–2 mol L–1

pH 5 [H3O+] = 10–5 mol L–1

Step 2 List the known and unknown information.

c1 = 10–2 mol L–1 V1 = 50 mL

c2 = 10–5 mol L–1 V2 = ?

Step 3 Find the new total volume of the solution. Using the stoichiometric relationship n = cV, we can say that c1V1 = c2V2, since in a dilution the number of moles is constant.

10 50 10

10 50

10

50 10

2 52

2

2

5

3

s � s

�s

� s

V

V

mL

Step 4 Calculate the volume of water that was added.

Wateradded 50 mL

49950 mL

49 95 L

�

�

�

50000

.

The ionic product of water

Pure water is neutral and has a pH of 7 at 25°C. There is a direct relationship between levels of H3O+ and OH– ions in aqueous solutions. At 25°C the product of their concentrations equals 1 s 10–14. This product is called the ionic product of water and is expressed by the following equation:

[H3O+] s [OH–] = 10–14(mol L–1)2 at 25°C.

For information about the pH

scale, go to the website for

this book and click on the pH

scale weblink (see Weblinks,

page 532).

For information about the pH

scale, go to the website for

this book and click on the pH

scale weblink (see Weblinks,

page 532).

Water molecules can react with themselves to produce a small number of H3O+ and OH– ions. The concentrations of these ions, when multiplied together, always equals 10–14 (mol L–1)2 (at 25°C). This is called the ionic product of water.

Water molecules can react with themselves to produce a small number of H3O+ and OH– ions. The concentrations of these ions, when multiplied together, always equals 10–14 (mol L–1)2 (at 25°C). This is called the ionic product of water.

Although water found in nature

is not often pure, the following

relationships are true at a

temperature of 25ºC.

In pure water and other neutral

solutions:

[H3O

+] = [OH

–] = 10

–7 pH = 7

In acidic solutions:

[H3O

+] > [OH

–]

[H3O

+] > 10

–7 pH < 7

In basic solutions:

[H3O

+] < [OH

–]

[H3O

+] < 10

–7 pH > 7

Although water found in nature

is not often pure, the following

relationships are true at a

temperature of 25ºC.

In pure water and other neutral

solutions:

[H3O

+] = [OH

–] = 10

–7 pH = 7

In acidic solutions:

[H3O

+] > [OH

–]

[H3O

+] > 10

–7 pH < 7

In basic solutions:

[H3O

+] < [OH

–]

[H3O

+] < 10

–7 pH > 7


In a neutral solution, such as water, [H3O+] = [OH–]. Substituting [H3O+] for [OH–], the above expression becomes:

[ ] [ ] ( )

[ ] (

H O H O mol L

H O

3 314 1 2

32 14

10

10

� �

�

s �

� mmol L

H O mol L

� �

1 2

37 110

)

[ ]

Therefore both [H3O+] and [OH–] are equal to 10–7 mol L–1 in a neutral solution at 25°C. It follows that if the amount of either [H3O+] or [OH–] in aqueous solution is increased, the other quantity must decrease by a corresponding factor. The equation for the ionic product of water has useful applications in pH calculations, due to the fact that pH is always measured in terms of [H3O+] or [H+]. When the concentration of a base is given, we need to convert this information to the concentration of H3O+, using the ionic product of water, before we can determine the pH.

Sample problem: 15.14 Using the ionic product to find the pH of a basic solution

The concentration of [OH–] in a cleaning product was found to be 10–3 mol L–1. Find the pH of the solution.

[ ] [ ] )

[ ] [ ]

H O OH (mol L

H O

13

14 2

33

10

10

�

�

s �

s �110

10

1010

14 2

3

14

311

�

< � �

(mol L

H O mo

1 )

[ ] ll L

pH

1

< �11

Litmus paper can be used to test for

acids (red), bases (blue) + neutrals

(yellow).

Although pH is usually in the

range of 1 to 14, it is not confined to

this range. Concentrated solutions

of strong acids may have a pH value

of less than zero. For example, in

a 10 mol L–1 HCl solution, [H3O+]

is equal to 101 mol L–1, giving a pH

of –1.

Concentrated solutions of strong

bases may have a pH value that

exceeds 14. For example, in a

10 mol L–1 NaOH solution, [OH–]

is 101 mol L–1. Therefore [H3O+] is

10–15 mol L–1, giving a pH of 15.

These calculations assume that

ionisation is complete.

Litmus paper can be used to test for

acids (red), bases (blue) + neutrals

(yellow).

Although pH is usually in the

range of 1 to 14, it is not confined to

this range. Concentrated solutions

of strong acids may have a pH value

of less than zero. For example, in

a 10 mol L–1 HCl solution, [H3O+]

is equal to 101 mol L–1, giving a pH

of –1.

Concentrated solutions of strong

bases may have a pH value that

exceeds 14. For example, in a

10 mol L–1 NaOH solution, [OH–]

is 101 mol L–1. Therefore [H3O+] is

10–15 mol L–1, giving a pH of 15.

These calculations assume that

ionisation is complete.

For any aqueous solution, [H3O+] s [OH–] will always equal 10–14 (mol L–1)2 at 25°C. This fact is especially useful when the [OH–] is given. The [H3O+], and hence the pH, can then be calculated.

For any aqueous solution, [H3O+] s [OH–] will always equal 10–14 (mol L–1)2 at 25°C. This fact is especially useful when the [OH–] is given. The [H3O+], and hence the pH, can then be calculated.


Revision questions

26. What is pH and how can it be calculated?

27. What is the difference between a dilute and a concentrated solution of an acid or base?

28. Determine the pH of the following given that [H3O+] is:(a) 10–5 mol L–1

(b) 0.1 mol L–1

29. Determine the pH of the strong acid HClO4 whose concentration is 0.001 mol L–1

30. Determine the pH of the following given that [OH–] is(a) 10–4 mol L–1

(b) 0.1 mol L–1

31. 90 mL of water is added to 10 mL of HNO3 with pH 2. What is the new pH?

Percentage yield

When an equation is used to calculate the amount of product that will form during a reaction, then a value for the theoretical yield is obtained. This is the maximum amount of product that could be formed from a given amount of reactant. By contrast, the amount of product that actually forms when the reaction is carried out in a laboratory is called the actual yield. The actual yield is often less than the theoretical yield. The percentage yield is the ratio of the actual yield to the theoretical yield. It measures the efficiency of the reaction.

percentage yieldactual yield

theoretical�

yields100%

A percentage yield should not be larger than 100%. Many factors may cause percentage yields to be less than 100%. These include the following:

• Reactions do not always go to completion.• Impure reactants and competing side reactions may cause other products to

be formed.• During purification, some of the product may be lost.

Copper ores usually contain only

about 2% copper. The effort and

cost of extracting copper from its

ores, however, is justified due to

the enormous usage of copper in

society.

Copper ores usually contain only

about 2% copper. The effort and

cost of extracting copper from its

ores, however, is justified due to

the enormous usage of copper in

society.


Sample problem: 15.15 Determining the percentage yield of a reaction

When copper is heated with an excess of sulfur, copper(I) sulfide is formed according to the equation:

2Cu(s) + S(s) Cu2S(s)

In a particular experiment, 2.40 g of copper was heated with excess sulfur to yield 2.85 g of copper(I) sulfide. What is the percentage yield of this reaction?

Solution:

Step 1 From the equation, identify the known and unknown quantities of substance.

The known quantity is the 2.85 g of copper(I) sulfide that formed from the initial 2.40 g copper sample.

Step 2 Calculate the number of moles of the known quantity of substance (copper) that was completely used up in the reaction.

n( ).

.Cu mol�

2 4063 6

Step 3 Find the molar ratio from the equation and use it to calculate the expected number of moles of the product, Cu2S.

n (Cu2S) : n (Cu) = 1 : 2

n( ).

.Cu S2

2 4063 6

12

� s

Step 4 Calculate the expected yield of the product, Cu2S.

expected mass of Cu2S �2.40

s1s 159.3

63.6 2� 3.01 g

Step 5 Calculate the percentage yield of Cu2S.

percentage yield of Cu2S �actual yield

expected yield

�

2.85s 100

3.01� 94.7%

Revision questions

32 Calculate the percentage yield when:(a) 2.34 g of aluminium reacts with an excess of copper(II) sulfate solution to

produce 3.89 g of copper, according to the equation: 2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)

(b) 63.5 g of iron(III) oxide reacts with an excess of carbon monoxide to produce 35.6 g of iron, according to the equation:

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

The yield of a reaction is the actual amount of product obtained compared to the theoretical amount that could have been obtained. It is often expressed as a percentage by using the formula:

percentage yield

actual yield

theoretical yi

�

eelds100

The yield of a reaction is the actual amount of product obtained compared to the theoretical amount that could have been obtained. It is often expressed as a percentage by using the formula:

percentage yield

actual yield

theoretical yi

�

eelds100


Chapter review

Summary

Stoichiometry is the study of the relative amounts — ratios — of substances involved in chemical reactions. This chapter introduced:

mass–mass stoichiometrymass–concentration stoichiometrystoichiometric problems involving solutionsacid–base titrationspH calculations.

The coeffi cients in a balanced chemical equation tell us the relative numbers of moles of reactants and products and can be used to fi nd the mole ratio of one substance to another.A four-step approach may be used to solve stoichiometric problems.Step 1 Write a balanced chemical equation for the

reaction, identifying the known (given) and unknown (required) quantities of substance.

Step 2 Calculate the number of moles of the known quantity of substance.

Step 3 From the equation, fi nd the mole ratio that states the proportion of known to unknown quantities in the reaction and use it to calculate the number of moles of the required substance.

Step 4 Calculate the required quantity of the substance.

Whenever two or more reactants are given in a chemical reaction, the limiting reactant must be identifi ed before we can calculate the maximum amount of product that may form. A limiting reactant is completely used up in a chemical reaction. The other reactants are known as excess reactants and are not used up.A solution is made up of a dissolved substance, called a solute, dissolved in a liquid, called the solvent.

The unit of molar concentration, or molarity, is defi ned as the number of moles per litre (mol L–1) of solution. This concept can also be used to calculate ionic concentrations.To prepare a particular volume of solution of known concentration, the following steps should

be followed:Step 1 Calculate the number of moles of solute

that are needed to obtain the correct concentration of solution for the volume of solvent to be used, according to the formula n = cV.

Step 2 Calculate the mass of the solute needed, using the formula m = nM.

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Step 3 Partially fi ll a volumetric fl ask with water, and add the correct mass of solute.

Step 4 Dissolve the solute.Step 5 Add water to the required volume.When a solution is diluted, the number of moles of solute remains the same. The concentration or volume of a diluted solution may be calculated according to the formula:

c1V1 = c2V2

In acid-base titrations, we can use stoichiometry to fi nd the relative amounts of acid or base required for neutralisation. The process involves:

an aliquot, or known volume, of the solution of unknown concentrationa standard solution of known concentration with which the solution of unknown concen tration is reacteda suitable indicator to ensure that the end point (where the solution changes colour), closely matches the equivalence point where chemically equivalent amounts of acid and base, according to the mole ratio of the equation, are present.

A titre is the volume of standard solution delivered from the burette during the titration.The pH of a solution may be calculated by the formula

pH = –log10[H3O+]

Also, from this formula, it follows that

[H3O+] = 10–pH

The ionic product of water is:

[H3O+] s [OH–] = 10–14 (mol L–1)2 at 25°C.

This is useful for calculating the pH of basic solutions. For a given [OH–], [H3O+] may be calculated, and hence the pH.The products of a chemical reaction may not always be formed in the predicted quantities.

The maximum amount of a product that can be formed from a given quantity of reactants is known as the theoretical yield.The actual yield is the amount of product that is obtained in a chemical reaction.The percentage yield of a chemical reaction is the actual yield expressed as a percentage of the theoretical yield, according to the formula:

percentage yieldactual yield

theoretical yi�

eelds100%

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Multiple choice questions

1. Magnesium reacts with oxygen according to the equation:

2Mg(s) + O2(g) 2MgO(s)

In this reaction:A 2 g of magnesium will react with 1 g of oxygen

to produce 2 g of magnesium oxideB 2 g of magnesium will react with exactly 1 g of

oxygenC 2 moles of magnesium are needed for every

mole of oxygen usedD 2 moles of oxygen are needed for every mole of

magnesium used. 2. Before digital printing, sodium thiosulfate, Na2S2O3,

known as ‘hypo’ by photographers, was used to remove excess silver bromide, AgBr, in the liquid-based film-developing process according to the equation:

2Na2S2O3(aq) + AgBr(s) Na3Ag(S2O3)2(aq) + NaBr(aq)

In the balanced equation for the reaction between sodium thiosulfate and silver bromide:A 1 mole of sodium thiosulfate solution will pro-

duce 1 mole of sodium bromide solutionB 3 moles of sodium thiosulfate solution will pro-

duce 2 moles of Na3Ag(S2O3)2

C 1 mole of sodium thiosulfate solution will pro-duce two moles of sodium bromide solution

D 1 mole of silver bromide will produce 1 mole of Na3Ag(S2O3)2.

3. The combustion of propane in air is represented by the equation:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

To produce 44 g of carbon dioxide in this reaction:A one-third of a mole of propane will react with

five-thirds of a mole of oxygenB 1 mole of propane will react with 1 mole of

oxygenC 1 mole of propane will react with 5 moles of

oxygenD 3 moles of propane will react with 15 moles of

oxygen. 4. The ore iron pyrites, FeS2, is used in a multi-step

process as a source of iron. In the first stage of this process, the ore is roasted to convert the iron(IV) sulfide to iron(III) oxide according to the equation:

4FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)

What is the maximum mass of iron(III) oxide which may be formed from the roasting of 60 tonnes of the ore? (1 tonne = 106g)A 20 tonnesB 40 tonnesC 44 tonnesD 80 tonnes

5. Phosphorus may be prepared from calcium phosphate according to the equation:

2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) P4(s) + 10CO(g) + 6CaSiO3(s)

How much phosphorus can be produced if 1000 kg of calcium phosphate is used completely?A 50 kgB 100 kgC 200 kgD 400 kg

6. The mass of potassium hydroxide, KOH, needed to produce 200 mL of a 0.25 mol L–1 solution is:A 0.050 gB 2.8 gC 50 gD 2.8 kg.

7. A solution containing 0.001 mole of HCl will react exactly with:A 10 mL of 0.1 mol L–1 NaOHB 10 mL of 0.01 mol L–1 NaOHC 1 mL of 0.1 mol L–1 NaOHD 1 mL of 0.01 mol L–1 NaOH.

8. A solution contains a mixture of two salts, potassium nitrate and potassium sulfate. If the concentration of potassium ions is 1.40 mol L–1 and the concentration of nitrate ions is 0.600 mol L–1, what is the concentration of the sulfate ions?A 0.400 mol L–1

B 0.800 mol L–1

C 0.900 mol L–1

D 1.800 mol L–1

9. A student dissolved 2.00 g of sodium hydroxide in 1.00 L of water to produce solution A. The student then diluted solution A with water to ten times its volume to produce solution B. Solution B was then diluted by the student with water to ten times its volume to produce solution C. What is the concentration of solution C?A 0.500 mol L–1

B 0.0500 mol L–1

C 0.00500 mol L–1

D 0.000500 mol L–1.10. In the titration of a strong acid against a strong base,

which of the following statements best describes the equivalence point?A the point at which the rate of the forward

reaction equals the rate of the reverse reactionB the point at which the same number of moles

of acid and base have been put into the flaskC the point at which the first sign of colour change

is seenD the point at which equal numbers of moles

of H+(aq) and OH–(aq) have been put into the flask


11. The end point in an acid–base titration is the point when:A the solution is neutralB the indicator changes colourC equal volumes of reactants have been mixedD reactants have been mixed in the appropriate

stoichiometric ratio.12. At 25°C, a solution is acidic if:

A [H3O+] = 10–14 mol L–1

B [OH–] = 10–7 mol L–1

C [H3O+] = 10–7 mol L–1

D [H3O+] = 10–1 mol L–1.13. The pH of a solution of 0.01 mol L–1 aqueous sodium

hydroxide is:A equal to 2B equal to 7C equal to 12D between 7 and 11.

14. In the combustion of a certain hydrocarbon, 16.0 g of CO2 is produced. This represents a 75% yield. What is the theoretical yield?A 12 gB 21.3 gC 32.0 gD 44.0 g

Review questions

Calculations using balanced equations

1. The reaction between hydrazine, N2H4, and hydrogen peroxide, H2O2, is used in some rocket engines. The balanced equation is:

N2H4(l) + 7H2O2(aq) 2HNO3(aq) + 8H2O(l)

If 1.5 mol of hydrazine was used, find:(a) the number of moles of H2O2 required(b) the number of moles of HNO3 and H2O

produced. 2. Balance the equations and then determine the

number of moles of each chemical which reacts or is produced.

(a) ______ N2(g) + ______ Cl2(g) ______ NCl3(l)

n (N2) n (Cl2) n (NCl3) (i) ______ 6.0 mol 4.0 mol (ii) 4.0 mol ______ ______ (iii) 3.2 mol ______ ______

(b) __________ C2H8(g) + __________ O2(g) __________ CO2(g) + _________ H2O(l)

n(C2H8) n(O2) n(CO2) n(H2O) (i) ______ 0.300 mol ______ ______ (ii) ______ ______ 4.00 mol ______ (iii) 6.00 mol ______ ______ ______

Mass–mass stoichiometry

3. Nitrogen in the cylinder of a car reacts with oxygen to produce the pollutant nitrogen monoxide.

N2(g) + O2(g) 2NO(g)

(a) How many moles of nitrogen are needed to produce 1.52 mol nitrogen monoxide?

(b) How many grams of oxygen are needed for this reaction?

4. Barbecues burning charcoal briquettes are unsafe for indoor use because of the colourless, odourless, poisonous gas produced.(a) Find the n(O2) gas that reacts with 3.5 g of

charcoal briquettes (assume pure C) to produce carbon monoxide.

(b) If there is a plentiful supply of air, a safe colourless, odourless gas is produced. Find the mass of this gas produced if the same amount of charcoal is burnt. You will need to write another equation.

5. Hydrochloric acid can be used to clean rust off iron and steel surfaces. The equation for the reaction is:

Fe(OH)3(s) + 3HCl(aq) FeCl3(aq) + 3H2O(l)

What mass of rust will react with 0.540 mol of HCl? 6. Milk of magnesia, Mg(OH)2, is used to neutralise

excess stomach acid, HCl, when a person has an upset stomach. What mass of magnesium hydroxide will neutralise 0.242 g of hydrochloric acid?

7. In respiration, the equation for the reaction that produces energy in our bodies is:

C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l)

If 8.90 g of glucose is used, determine:(a) the mass of oxygen needed(b) the mass of carbon dioxide produced.

8. When sodium and chlorine are reacted, table salt is produced.(a) Write the balanced equation for the reaction.(b) If 10.0 g of sodium were used, how much table

salt would be produced? 9. Aspirin, C9H8O4, is taken by many people for the

relief of headaches or minor pain. It is produced from salicylic acid, C7H6O3, and acetic anhydride, C4H6O3. The equation is:

C7H6O3 + C4H6O3 C9H8O4 + C2H4O2

Find the mass of salicylic acid needed to produce 155 kg aspirin.

10. The silver used in jewellery and tableware becomes tarnished when exposed to air containing small amounts of hydrogen sulfide. The tarnish is a layer of silver sulfide.

4Ag(s) + 2H2S(g) + O2(g) 2Ag2S(s) + 2H2O(g)

Calculate the mass of the tarnish when 0.025 g of silver is reacted.


11. Common fuels used to produce rocket thrust in order to put satellites into orbit are liquefied hydrogen and oxygen. What mass of hydrogen will combine with 2.03 kg of oxygen?

12. In the final step of the Contact Process, where sulfuric acid is manufactured industrially, controlled addition of water to oleum, H2S2O7, produces pure sulfuric acid. The equation for the reaction is:

H2S2O7(l) + H2O(l) 2H2SO4(l)

Calculate the mass of sulfuric acid produced when 5.00 kg of oleum is dissolved in the water.

13. The metal tungsten, used to make the filament for incandescent light bulbs, can be obtained from its oxide by reduction with hydrogen.

WO3(s) + 3H2(g) W(s) + 3H2O(g)

(a) What mass of tungsten can be obtained from 200 g of its oxide?

(b) What mass of hydrogen is required?14. The thick haze commonly seen in highly industrial

areas is caused by ammonium sulfate, (NH4)2SO4. This compound results when ammonia, NH3, gas reacts with droplets of sulfuric acid.(a) Write the balanced equation for this reaction.(b) Calculate the mass of ammonium sulfate

formed from the complete reaction of 700 g of ammonia.

15. In a class experiment, a strip of copper was placed in an aqueous solution of silver nitrate.

Initial mass of copper strip = 4.36 g Final mass of copper strip = 2.21 g

(a) Write an equation for the reaction.(b) What mass of silver formed in the reaction?

16. Calcium carbonate can be decomposed by heating:

CaCO3(s) CaO(s) + CO2(g)

When 24.8 g of calcium carbonate was heated, 13.1 g of calcium oxide was obtained.(a) Calculate the amount of calcium oxide which

theoretically should have been obtained.(b) Calculate the percentage yield of calcium

oxide.17. It has been estimated that 4.0 s 109 tonnes of oxygen

are produced annually by photosynthesis in plants to produce glucose.

6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)

(a) What mass of carbon dioxide is consumed in producing this amount of oxygen?

(b) What mass of glucose is formed?18. A 1.50 g sample of paint pigment was dissolved and

the lead ions separated by precipitation as solid PbSO4. The precipitate had a mass of 0.0806 g. Find the percentage of lead in the pigment.

Limiting reagent calculations

19. When sodium is placed on water it reacts violently to produce hydrogen gas and a solution of sodium hydroxide.(a) Write a balanced chemical equation for this

reaction.(b) If 9.2 g of sodium reacts completely, calculate

the mass of gas that would be produced.(c) What mass of water would react completely

with 4.6 g of sodium?(d) If 1.0 kg of sodium was placed in 300 mL of

water, which reactant would be used up first?20. Sodium and chlorine react according to the

following equation:

2Na(s) + Cl2(g) 2NaCl(s)

If 12.5 g of sodium and 25.5 g of chlorine are available for reaction, determine:(a) the limiting reactant(b) the mass of NaCl formed.

21. Sulfur dioxide gas escaping from a natural gas well may react with water to produce an acid which damages the environment when it falls to the ground as acid rain. If 32 g of sulfur dioxide reacts with 18 g of water:(a) how much H2SO3 is produced?(b) Which reactant is in excess and by how much?

22. Hydrazine is used in rocket fuel. It reacts with oxygen as follows:

N2H4(l) + O2(g) N2(g) + 2H2O(l)

In a particular rocket engine, 2.29 g of hydrazine and 3.14 g of oxygen are available for reaction.(a) Find the limiting reactant.(b) Determine which reactant is in excess and by

how many grams.(c) Calculate the mass of water produced.

23. Magnesium and chlorine react according to the following equation:

Mg(s) + Cl2(g) MgCl2(s)

If 20 g of magnesium and 20 g of chlorine are available for reaction, determine:(a) the limiting reactant(b) the mass of MgCl2 formed.

Concentration of solutions

24. Calculate the mass of solute in each of the following solutions:(a) 300 mL of 1.5 mol L–1 NaOH(b) 250 mL of 2.0 mol L–1 H2SO4

(c) 17.5 L of 1.5 mol L–1 Na2CO3

(d) 200 mL of 2.5 mol L–1 CuSO4·5H2O(e) 120 mL of 1.7 mol L–1 Mg(NO3)2.


25. Calculate the concentration (molarity) of solute in each of the following solutions:(a) 250 mL of solution containing 17 g sodium

bromide, NaBr(b) 500 mL of solution containing 200 g magnesium

sulfate, MgSO4

(c) 1.5 L of solution containing 1500 g magnesium nitrate, Mg(NO3)2

(d) 100 mL of solution containing 1000 g sodium phosphate, Na3PO4

(e) 2.00 L of solution containing 500 g potassium carbonate, K2CO3

(f) 40.0 mL of solution containing 12.5 g aluminium chloride, AlCl3.

26. Calculate the concentration of the following solutions:(a) 4.0 g of sodium hydroxide in 200 mL of

solution(b) 12.6 g of sodium carbonate in 350 mL of

solution(c) 5.35 g of magnesium carbonate in 500 mL of

solution.27. Calculate the number of moles of solute needed to

prepare:(a) 30 mL of 0.10 mol L–1 AgNO3

(b) 300 mL of 1.5 mol L–1 Mg(NO3)2

(c) 230 mL of 0.40 mol L–1 KCl(d) 2.5 L of 0.2 mol L–1 KNO3.

28. Calculate the mass of solute needed to make each of the following quantities of solution:(a) 250 mL of a 1.5 mol L–1 sodium bromide, NaBr,

solution(b) 500 mL of a 1.75 mol L–1 magnesium sulfate,

MgSO4, solution(c) 1.50 L of a 0.575 mol L–1 magnesium nitrate,

Mg(NO3)2, solution(d) 100 mL of a 0.850 mol L–1 sodium phosphate,

Na3PO4, solution(e) 2.00 L of a 0.005 mol L–1 potassium carbonate,

K2CO3, solution(f) 40.0 mL of a 2.30 mol L–1 aluminium chloride,

AlCl3, solution.29. How many grams of CH3COONa are obtained when

400.0 mL of a 0.500 mol L–1 solution of CH3COONa is evaporated to dryness?

30. What is the concentration of each of the following solutions?(a) 58.5 g of H2SO4 dissolved in enough water to

produce 2.00 L of solution(b) 2.7 g of HCl dissolved in enough water to

produce 500 mL of solution(c) 4.04 g of KNO3 dissolved in enough water to

produce 150.0 mL of solution(d) 234 g of sodium chloride dissolved in enough

water to produce 6.00 L of solution.

Concentration of ions in solution

31. Most common fertilisers contain nitrogen compounds. Ammonium nitrate, NH4NO3, is soluble in water and hence it is quickly taken up by the plant’s root system. If a 1.5 L bucket contains 150 g of ammonium nitrate calculate:(a) the concentration of the solution(b) the concentration of each ion in the solution.

32. Aluminium chloride, AlCl3, is found in anti-perspirants. A student wishes to do some tests on antiperspirant, so a jar of it is dissolved into a beaker containing 700 mL of water. The label on the jar states that it contains 200 g of aluminium chloride. Calculate:(a) the concentration of the solution(b) the concentration of each ion in the solution.

33. Calcium hydroxide, Ca(OH)2, is found in plaster and cement, and is used in the treatment of drinking water. If the concentration of OH– of a solution of calcium hydroxide is 0.050 mol L–1, calculate the concentration of the calcium hydroxide.

34. In an Fe2(SO4)3 solution the concentration of SO42–

is 0.25 mol L–1. What is the concentration of solute?

Dilution

35. Calculate the volume of stock solution needed to prepare the following:(a) 500 mL of a 0.750 mol L–1 solution of sulfuric

acid (sulfuric acid stock solution = 18 mol L–1)(b) 200 mL of a 2.5 mol L–1 solution of ammonium

hydroxide (ammonium hydroxide stock solution = 15 mol L–1)

(c) 350 mL of a 0.150 mol L–1 solution of acetic acid (acetic acid stock solution = 18 mol L–1).

36. Calculate the concentration (molarity) of:(a) 100 mL of 0.1 mol L–1 solution after dilution

with 10 mL water(b) 16 mL of 0.20 mol L–1 solution after dilution

with 4.0 mL water(c) 750 mL of 1.50 mol L–1 solution after dilution

with 150 mL water.37. What volume of water must be added to:

(a) 100 mL of 15 mol L–1 H2SO4 to dilute it to 5.0 mol L–1

(b) 130 mL of 3.50 mol L–1 HNO3 to dilute it to 1.00 mol L–1

(c) 170 mL of 2.60 mol L–1 HCl to dilute it to 0.250 mol L–1?

38. A student is required to make up 500 mL of 0.1 mol L–1 H2SO4(aq) by diluting a 0.5 mol L–1 solution of the acid. What volume of 0.5 mol L–1 H2SO4(aq) is required?

39. What volume of water is needed to produce a 2.00 mol L–1 NaCl solution if 2.35 g of NaCl are present?


40. A 3.50 g mass of MgCl2 is dissolved in 200 mL of water.(a) What is the concentration of the solution?(b) If 150 mL is added to the solution, what is the

concentration of the new, diluted solution?

Mass–concentration stoichiometry

41. What volume of 1.3 mol L–1 HCl will just neutralise 2.5 g of Ca(OH)2?

Ca(OH)2(s) + 2HCl(aq) CaCl2(aq) + 2H2O(l)

42. What mass of copper is required to react completely with 250 mL of 0.100 mol L–1 AgNO3?

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

43. Chlorine gas was bubbled through 120 mL of 0.300 mol L–1 NaBr until all the bromide ions were replaced. How many moles of chlorine gas reacted?

Cl2(g) + 2NaBr(aq) 2NaCl(aq) + Br2 (aq)

44. 25.0 mL of arsenic(III) chloride solution is reacted with an excess of 0.200 mol L–1 NaOH solution, forming 3.15 g of precipitate of arsenic hydroxide.

AsCl3(aq) + 3NaOH(aq) As(OH)3(s) + 3NaCl(aq)

(a) Find the number of moles of arsenic hydroxide.

(b) Find the number of moles of sodium hydroxide that reacted.

(c) Find the concentration of the original arsenic chloride solution.

(d) Find the minimum volume of 0.200 mol L–1 NaOH required to react with the 25.0 mL AsCl3.

45. A sample of water was known to contain chloride ions. To determine the concentration of chloride ions in the water sample, a solution of silver nitrate was added to 500 mL of the water. 1.26 g of silver chloride precipitate formed. What was the concentration of chloride ions in the water sample?

46. What volume of water, in mL, must be added to 400 mL of a solution containing 32.0 g of KBr to form a 0.100 mol L–1 solution of KBr?

47. What mass of Cr(OH)3 would be produced if 400 mL of 0.250 mol L–1 KOH was added to a solution containing excess CrCl3? The equation for the reaction is:

3KOH(aq) + CrCl3(aq) Cr(OH)3(s) + 3KCl(aq)

48. The equation for the reaction between barium hydroxide and aluminium nitrate is given below:

3Ba(OH)2(aq) + 2Al(NO3)3(aq) 2Al(OH)3(aq) + 3Ba(NO3)2(aq)

What volume of 1.55 mol L–1 Ba(OH)2 is required to produce 287 g of Al(OH)3?

Solution stoichiometry

49. Sodium chloride reacts with silver nitrate to give a white precipitate of silver chloride.(a) Write an equation for the reaction.(b) What volume of 2.0 mol L–1 silver nitrate

will react with 120 mL of 1.5 mol L–1 sodium chloride?

50. 50 mL of 1.2 mol L–1 sulfuric acid was added to 30 mL of sodium hydroxide in order to neutralise it. Find the molarity of the sodium hydroxide.

51. An iron ore sample may be analysed by preparing an iron(II) ion solution of the ore which is then reacted with acidic permanganate solution as in the given equation. If 15.0 mL of 0.0100 mol L–1 MnO4

– solution is required to completely react with 10.0 mL of the Fe2+ solution, determine the concentration of the iron(II) ion solution.

MnO4–(aq) + 8H+(aq) + 5Fe2+(aq)

5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

52. What volume of a 2.30 mol L–1 solution of Na3PO4 is required to react completely with 560 mL of a 3.25 mol L–1 solution of Ca(ClO3)2, according to the reaction:

3Ca(ClO3)2(aq) + 2Na3PO4(aq) Ca3(PO4)2(aq) + 6NaClO3(aq)

53. What volume of 0.250 mol L–1 HNO3 will react completely with 280 mL of 0.200 mol L–1 H2S according to the reaction:

2HNO3(aq) + 3H2S(aq) 3S(s) + 4H2O(l) + 2NO(g)

Acid–base titrations

54. The concentration of vinegar was determined by titration. 20.00 mL of vinegar required 43.20 mL of 0.350 mol L–1 sodium hydroxide according to the following equation. Find the concentration of vinegar.

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

55. What volume of 0.460 mol L–1 H2SO4 is required to neutralise 24.00 mL of 0.620 mol L–1 NaOH?

56. 1.365 g of anhydrous sodium carbonate, Na2CO3, is dissolved in 250.0 mL of water. 20.00 mL aliquots of this solution were titrated with nitric acid. An average titre of 21.95 mL was needed. Find the concentration of the nitric acid.

pH calculations

57. The concentration in water of a monoprotic acid with the formula HX is 0.10 mol L–1. The con-centration of the H3O+ ion in this solution is 0.010 mol L–1.


(a) What is a monoprotic acid?(b) Write an equation to illustrate the ionisation

reaction of HX in water.(c) Is HX a strong or weak acid? Explain.(d) Would a solution of HX be a good conductor of

electricity? Justify your response.58. Find the pH of solutions with:

(a) [H+] = 10–3 mol L–1 (b) [OH–] = 10–5 mol L–1.

59. Find the pH of the solution given that [H3O+] is:(a) 10–4 mol L–1

(b) 10–6 mol L–1

(c) 10–9 mol L–1

(d) 0.001 mol L–1

(e) 1 mol L–1.60. Find the pH of the solution given that [OH–] is:

(a) 10–14 mol L–1

(b) 10–10 mol L–1

(c) 10–1 mol L–1

(d) 0.01 mol L–1

(e) 1 mol L–1.61. Find the [H3O+] if the pH of the solution is:

(a) 1(b) 2(c) 3(d) 10.

62. Find the [OH–] if the pH of the solution is:(a) 3(b) 12(c) 7(d) 1.

63. Find the pH of the solution if:(a) [H3O+] = 100 mol L–1

(b) [OH–] = 10–12 mol L–1

(c) [H3O+] = 10–2 mol L–1

(d) [H3O+] = 0.001 mol L–1

(e) [OH–] = 0.001 mol L–1.

64. Find the pH of:(a) an aqueous solution containing 0.073 g HCl in

2.00 L water(b) an aqueous solution containing 0.80 g NaOH in

2.00 L water.65. The pH of blood is 7.4.

(a) Is this acidic or basic?(b) Calculate the hydrogen ion concentration.

66. What volume of water is required to dilute 30 mL of a nitric acid solution of pH 3 to pH 5?

67. 50 mL of a solution of pH 2 is added to 450 mL of water. What is the new pH?

Percentage yield

68. Why may a reaction that has a percentage yield of 95% be described as inefficient?

69. Calcium carbonate can be decomposed by heating according to the equation below. What is the percentage yield of calcium oxide in this reaction if 12.6 g of CaO is obtained when 25.9 g of CaCO3 is heated?

CaCO3(s) CaO(s) + CO2(g)

70. Calculate the percentage yield of iron in a reaction where 67.3 g of iron(III) oxide reacts with an excess of carbon monoxide to produce 41.8 g of iron, according to the equation:

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

71. When 54.8 g of silicon dioxide is heated with an excess of carbon, 32.5 g of silicon carbide is produced in addition to carbon monoxide gas. What is the percentage yield of this reaction?

72. What is the percentage yield of copper in a reaction which produces 2.58 g of copper from 1.56 g of aluminium and an excess of copper(II) sulfate according to the equation:

2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)


Exam practice questions

In a chemistry examination you will be required to answer a number of short and extended response

questions. Your answers will be assessed on how well you:

• use your knowledge and the information provided

• communicate using relevant chemistry terminology and concepts

• present a logical, well-structured answer to the question.

EXTENDED RESPONSE QUESTIONS

1. Aspirin is known chemically as acetylsalicylic acid and has the formula C9H8O4. It can be prepared from ethanoic acid and salicylic acid according to the equation:

C7H6O3(s) + C4H6O3(l) C9H8O4(s) + C2H4O2(l)(salicylic acid)

Calculate the mass of salicylic acid that would be required to produce 10.0 g of aspirin. 5 marks

2. Anhydrous sodium carbonate is often used to make standard solutions of sodium carbonate. This solution may then be used in acid-base titrations to calculate the exact concentration of acidic solutions.

(a) Calculate the mass of anhydrous sodium carbonate (Na2CO3) that is required to make 500.00 mL of 0.0500 mol L–1 solution. 3 marks

(b) Exactly 25.00 mL of the solution from (a) is reacted with hydrochloric acid. The equation for this reaction is:

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + CO2(g) + H2O(l)

It is found that 23.50 mL of the acid is required for exact neutralisation. Calculate the molarity of the hydrochloric acid solution. 4 marks

numerical connections —...

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