1.3 Chemical equations

1.3.1 Deduce chemical equations when all reactants and products are given.

Reactant 1 + Reactant 2 Product

A Chemical Reaction is basically a reaction between reactants to form products. A chemical reaction is

represented either with the chemical formulas or the symbols involved in the reaction.

Let’s consider the combustion of Methane.

Chemical Equation

Symbolic Equation

However, the symbolic equation above is slightly incorrect. If you translate the symbolic equation to the equation with

words, you still get “Methane + Oxygen –> Hydrogen + Carbon Dioxide.” Then why is it wrong?

This is because the symbolic equation does not follow the Law of Conservation. Law of conservation is similar to

Conservation of Energy, where energy is conserved. Basically, here what starts in the beginning must last till to the end.

Speaking more formally, this means that each side of both the reactants and products must have the same number of

every type of atom. AKA we need to BALANCE

The original equation is:

Right now, there’s an imbalance of both oxygen and hydrogen atoms. There are four hydrogen atoms on the left, and two

on the right.

To remedy this, why don’t I first tag a 2 in front of the H2O on the products side?

The equation now becomes:

Since I added a two in front of H2O, I’m multiplying 2 to the Oxygen atom in H2O as well.

Now, there are 2 oxygen atoms on the left, and 4 on the right!

I’ll change the coefficient of the oxygen on the reactant side to 2, so I will have 2 x (2 oxygen atoms) = 4 oxygen

atoms on the reactants side.

Now, all the atoms are balanced!

Reactants

C = 1

H = 4

O = 4

Products

C= 1

H =4

O= 4

Now they are balanced!

1.3.2 Identify the mole ratio of any two species in a chemical equation.

This is actually quite easy. To identify mole ratio between two species in a chemical equation, simply find the ratio

between the coefficients!

So for the previous reaction:

We can actually write it like this:

If the coefficient = 1, you don’t really have to write it in front of the substance, but I am doing this just to make the point

about mole ratio identification.

What is the mole ratio between the Carbon Dioxide and Oxygen?

Since the coefficient of methane is 1, and 2 for oxygen, the mole ratio will be 1:2. Basically what this means is that 1 mole

of methane reactants with 2 moles of oxygen.

1.3.3 Apply the state symbols (s), (l), (g) and (aq).

We can add state symbols next to the substances in a symbolic equation.

(s) = solid. (l) = liquid. (g) = gas (aq) = aqueous (relating to or dissolved in water)

Example application:

1.4 Mass and Gaseous volume relationships in chemical reactions

1.4.1 Calculate theoretical yields from chemical equations.

The steps for calculating Masses of Reactants and products in Chemical Reactions:

Example: What mass of CO2 can be absorbed by 1.00kg of LiOH

LiOH + Co2 Li2CO3 + H2O

1. Balance the equation for the reaction.

2 LiOH + Co2 à Li2CO3 + H2O

2. Convert the known mass of the reactant (or product) to moles.

1000g (from 1.00kg) x 1 mol

23.95g (AMU of LiOH) = 41.8 moles

3. Use the balanced equation to set up the appropriate mole ratios.

1 CO2

2 LiOH

4. Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product.

1 Co2 x 41.8 mol LiOH = 20.9 mol CO2

2 LiOH

5. Convert from moles back to grams (if the question asks for mass)

20.9 mol x 44g CO2 (AMU)

1mol CO2 = 919.6 g = 920 g

1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.

LIMITING REACTANTS VS EXCESS REACTANTS

Limiting decides how much of the amount of product that can be formed.

Excess is when there’s more than enough of a substance to react with the limiting reagent (reactant).

The steps for Solving limiting or excess reactant and products:

Example: If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO. Find the limiting reactant and how many

grams of N2 will be formed. (In other words how much much N2 can we get from NH3 and CuO)

NH3 + CuO N2 + 3Cu + H2O

1. Balance the equation for the reaction.

2 NH3 + 3CuO N2 + 3Cu + 3H2O

2. Convert the know masses to moles.

18.1.g NH3 x 1 mol = 1.06 moles

17.034 g

90.4g CuO x 1 mol = 1.14 moles

79.55 g

3. Determine which reactant is limiting (the other must be excess). Choose any of the two know masses to see which has

too much or too little.

1.06 mol NH3 x 3 mol CuO = 1.59 mol CuO ---- This is the number of moles needed.

2 mol NH3 , We only have 1.14 moles, therefore

CuO is the limited reactant.

4. Using the amount of the limiting reactant and the appropriate mole ratios, compare the number of moles of the desired

product. (This step can be skipped since it is just the proof that step 3 is true.)

Mol CuO = 3 = 1.5

Mol NH3 = 2

Mol CuO = 1.14 = 1.08

Mol NH3 =1.06

5. Convert from moles to grams using the molar mass.

First find the moles. 1.14 mol x 1mol N2 = 0.38 mol N2

3mol CuO

Then the moles to grams= 0.38 mol N2 x 28.0g N2 = 10.6 N2

1.4.3 Solve problems including theoretical, experimental and percentage yield.

Limiting Reagents and Percentage Yield

"If one reactant is entirely used up before any of the other reactants, then that reactant limits the maximum yield of the product."

Problems of this type are done in exactly the same way as the previous examples, except that a decision is made before the ratio comparison is done. The decision that is made is "What reactant is there the least of?"

Example Problem #1

Methane, CH4, burns in oxygen to give carbon dioxide and water according to the following equation: CH4 + 2 O2 ------> CO2 + 2 H2O

In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel vessel. Find the limiting reactant, if any, and calculate the theoretical yield, (in moles) of water.

Solution:

In any limiting reactant question, the decision can be stated in two ways. Do it once to get an answer, then do it again the second way to get a confirmation.

According to the equation: 1 mol CH4 = 2 mol O2

If we use up all the methane then: 1 mol CH4 = 2 mol O2 0.25 mol x x = 0.50 mol of O2 would be needed.

We have 1.25 mol of O2 on hand. Therefore we have 0.75 mol of O2 in excess of what we need. If the oxygen in is excess, then the methane is the limiting reactant.

Confirmation: If we use up all the oxygen then

1 mol CH4 = 2 mol O2 x 1.25 mol x = 0.625 mol of methane.

We don't have 0.625 moles of methane. We have only 0.25 moles. Therefore the methane will be used up before all the oxygen is. Again the methane is the limiting reactant.

We now use the limiting reactant to make the mole comparison across the bridge to find the amount of water produced.

1 mol CH4 = 2 H2O 0.25 mol x x = 0.50 mol of H2O would be produced.

Conclusion: When 0.25 mole of methane and 1.25 mole of oxygen are mixed and reacted according to the equation, the methane is the limiting reactant and the maximum yield of water will be 0.50 moles.

Example Problem #2

Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed. Which is the limiting reactant? What is the maximum yield of CCl4 in moles and in grams?

Solution: Start with the equation: CHCl3 + Cl2 -------> CCl4 + HCl Did you check to see if it was balanced?

Calculate the molecular masses of the species needed in the problem. CHCl3 = 1 C = 1(12.01) = 12.01 Cl2 = 2 (35.45) = 70.90 g/mol 1 H = 1(1.01) = 1.01 H2O = 2 H = 2 (1.01) = 2.02 3 Cl = 3(35.45) = 106.35 1 O = 1 (16.00) = 16.00 119.37 g/mol 18.02 g/mol

Then calculate the moles of each of the reactants to be used. moles of CHCl3 = g = 25.00 g = 0.21 moles of CHCl3 are present. mm 119.37 g/mol

moles of Cl2 = g = 25.00 g = 0.35 moles of chlorine are present. mm 70.90 g/mol

Decision time. Which of the two reactants do you have the least of? From the balanced equation you can see that the chloroform and chlorine reactant in a one to one ratio. If we use all the chloroform then we get the following equation.

1 CHCl3 = 1 Cl2 0.21 mol x x = 0.21 moles of chlorine are needed.

We need 0.21 moles of chlorine. We have 0.35 moles of chlorine. Therefore chlorine is in excess. The chloroform must be the limiting reactant.

Confirmation: IF we use all the chlorine then:

1 CHCl3 = 1 Cl2 x 0.35 mol x = 0.35 moles of chloroform are needed.

If we use all the chlorine then we need 0.35 moles of chloroform. We have only 0.21 moles of chloroform. It is the reactant that we will run out of first. Therefore it is the limiting reactant.

Use the limiting reactant to cross the ratio bridge and find the number of moles of water made.

1 CHCl3 = 2 H2O 0.21 mol x x = 0.42 moles of H2O will be made.

Calculate the grams of water produced. grams = moles * molecular mass = 0.42 mol * 18.02 g/mol = 7.57 grams of water

Conclusion: When 25 grams of each reactant are mixed according to the equation, the chloroform is the limiting reagent and the maximum yield of water will be 0.42 moles or 7.57 grams.

Example Problem #3

Aluminum chloride, AlCl3, can be made by the reaction of aluminum with chlorine according to the following equation: 2 Al + 3 Cl2 ------> 2 AlCl3

What is the limiting reactant if 20.0 grams of Al and 30.0 grams of Cl2 are used, and how much AlCl3 can theoretically form?

Have you checked to make sure the equation is balanced correctly?

Find the molecular masses of all species involved. Al = 26.98 g/mol Cl2 = 70.90 g/mol AlCl3 = 133.33 g/mol

Convert the grams into moles. moles of Al = g/mm = 20.00 g/26.98 g/mol = 0.74 moles of aluminum on hand.

moles of Cl2 = g/mm = 30.00 g/70.90 g/mol = 0.42 moles of chlorine on hand.

Decision time: Which is the limiting reagent? IF we use all aluminum then:

2 Al = 3 Cl2 0.74 mol x x = 1.11 moles of chlorine are needed.

We don't have 1.11 moles of chlorine. We have 0.42 moles of chlorine. Therefore we will run out of chlorine first. It is the limiting reactant.

Confirmation: If we use all the chlorine then:

2 Al = 3 Cl2 x 0.42 mol x = 0.28 moles of aluminum are needed.

We have 0.74 moles of aluminum, therefore it is in excess. If it is in excess then the chlorine is the limiting reactant.

Use the limiting reactant to cross the ratio bridge and find the moles of AlCl3 that will be produced. 3 Cl2 = 2 AlCl3 0.42 mol x x = 0.28 moles of AlCl3 are produced

Grams of aluminum chloride are found with g = n * mm = 0.28 mol * 133.33 g/mol = 37.33 g

Conclusion: When 20.0 grams of aluminum and 30.0 grams of chlorine are reacted according to the above equation, the chlorine is the limiting reactant and the maximum yield of aluminum chloride is 0.28 moles or 37.33 grams.

1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.

Molar volume - volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure. There are two standards:

STP (standard temperature and pressure) = is 0 ºC and 1 atmosphere.

RTP (room temperature and pressure) = 20 ºC and 1 atmosphere.

Molar Volume

Avogadro’s Law states that: 1 mole of every gas occupies the same volume, at the same temperature and pressure.

At STP (standard temperature and pressure), volume =22.4 liters

At RTP (room temperature and pressure), this volume= 24 dm3 (liters)

We can also say:

The molar volume of a gas is 22.4 liters at STP (standard temperature and pressure). The molar volume of gas is 24 dm

3 at RTP (room temperature and pressure).

How to find the molar volume of a gas using the ideal gas law. The most common molar volume is the molar volume of an ideal gas at standard temperature and pressure (273 K and 1.00 atm)

Gas volumes from moles to gas Example: Calculate the volume of carbon dioxide gas, CO2, occupied by (a) 5 moles and (b) 0.5 moles of the gas occupied at STP. Solution: a) Volume of CO2 = number of moles of CO2 × 22.4 L = 5 × 22.4 = 112 L b) Volume of CO2 = number of moles of CO2 × 22.4 L = 0.5 × 22.4 = 11.2 L

Moles from gas volumes

Example: Calculate the number of moles of ammonia gas, NH3, in a volume of 80 L of the gas measured at STP. Solution: Volume of gas = number of moles × 22.414 L/mol

Gas volume from equations

From the equation for a reaction, we can tell how many moles of a gas take part. Using Avogadro’s Law, we can also work out its volume. Example: What volume of hydrogen will react with 22.4 liters of oxygen to form water? (All volumes are measured at STP) Solution: Step 1: Write a balanced equation for the reaction. 2H2 (g) + O2 (g) → 2H2O (l) Step 2: Calculate the volume. From the equation, 2 volumes of hydrogen react with 1 of oxygen or 2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen. The volume of hydrogen that will react is 44.8 liters.

Example: When sulfur burns in air it forms sulfur dioxide. What volume of this gas is produced when 1 g of sulfur burns? (Ar : S = 32) (All volumes are measured at STP) Solution: Step 1: Write a balanced equation for the reaction. S (s) + O2 (g) → SO2 (g) Step 2: Get the number of moles from the grams. 32 g of sulfur atoms = 1 mole of sulfur atoms So, 1 g = 1 ÷ 32 mole or 0.03125 moles of sulfur atoms 1 mole of sulfur atoms gives 1 mole of sulfur dioxide molecules So, 0.03125 moles of sulfur atoms gives 0.03125 moles of sulfur dioxide Step 3: Get the volume. 1 mole of sulfur dioxide molecules has a volume of 22.4 at STP So, 0,03125 moles has a volume of 0.03125 × 22.4 = 0.7 liters at STP So, 0.7 liters of sulfur dioxide are produced.

1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations.

1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal

gas

1.4.7 Solve problems using the ideal gas equation, PV = nRT

The Ideal Gas Equation

Before we look at the Ideal Gas Equation, let us state the four gas variables and one constant for a better understanding.

The four gas variables are: pressure (P), volume (V), number of mole of gas (n), and temperature (T). Lastly, the constant

in the equation shown below is R, known as the the gas constant, which will be discussed in depth further later.

equation is:

Another way to describe an ideal gas is to describe it in mathematically. Consider the following equation:

An ideal gas will always equal 1 when plugged into this equation. The greater it deviates from the number 1, the more it

will behave like a real gas rather than an ideal.

A few things should always be kept in mind when working with this equation, as you may find it extremely helpful when

checking your answer after working out a gas problem.

Pressure is directly proportional to number of molecule and temperature. (Since P is on the opposite side of the equation

to n and T)

Pressure, however, is indirectly proportional to volume. (Since P is on the same side of the equation with V)

Standard Temperature and Pressure (STP)

Standard condition of temperature and pressure is known as STP. Two things you should know about this is listed below.

The universal value of STP is 1atm (pressure) and 0o C degree. Note that

this form specifically stated 0o C degree, not 273 Kelvin, even though you

will have to convert into Kelvin when plugging this value into the Ideal Gas equation or any of the simple gas equations.

In STP, 1 mole of gas will take up 22.4 L of the volume of the container. Units of P, V and T

Factor Variable Units

Pressure P atm Torr Pa

mmHg

Volume V L m³

Moles n mol

Temperature T K

Gas Constant R* see Values of R table

Ideal Gas Law IB Units

Pressure atm Volume liters

n moles R L atm mol

-1 K

-1

Temperature Kelvin

Formulas to remember:

Example 1: What mass of argon is contained in an 18.6L container at 20°C if the pressure is 2.35 atm?

Example 2: What is the density of carbon tetrafluoride at 1.00 atm and 50 ºC?

Example 3: 5.0 g of neon is at 256 mm Hg and at a temperature of 35º C. What is the volume?

Step 1: Write down your given information: P = 256 mmHg V = ? m = 5.0 g R = 0.0820574 L·atm·mol-1K-1 T = 35º C Step 2: Convert as necessary:

Pressure:

Moles:

Temperature:

Step 3: Plug in the variables into the appropriate equation.

2.) What is a gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm? Step 1: Write down your given information: P = 143.5 atm V= 25 L n = 203 mol R = 0.0820574 L·atm·mol

-1 K

-1

T = ? You can skip Step 2 because all units are the appropriate units. Step 3: Plug in the variables into the appropriate equation.

Step 4: You are not done. Be sure to read the problem carefully, and answer what they are asking for. In this case, they are asking for temperature in Celsius, so you will need to convert it from K, the units you have.

1.4.8 Analyse graphs relating to the ideal gas equation.