unit 9 stoichiometry

Unit 9 Stoichiometry

Mole-Mole, Mass-Mass, Mass-Volume, Volume-Volume, Limiting Reactant and Percent Yield

Stoichiometry

The study of quantitative, or measurable relationships that exist in chemical formulas and chemical reactions.– One form of stoich involves converting among

moles, particles, mass and volume of chemical formulas. (Unit 8)

– The other form of stoich is concerned with chemical reactions and involves the relationships between reactants and products in a chemical reaction. (Unit 9)

Mole-Mole Problems

The coefficients in a balanced chemical equation can be interpreted both as the number of particles and as the number of moles involved in the reaction.

Observe the chemical equation below.

2H2 + O2 2H2OAccording to the equation: 2 moles of hydrogen and

one mole of oxygen produce 2 moles of water.

Mole-Mole Problems

Suppose you wanted to know how many moles of hydrogen and oxygen are needed to produce 5.48 moles of water.

First determine the molar ratios indicated by the equation.

2H2 + O2 2H2OH2 to H2O is 2:2O2 to H2O is 1:2

Mole-Mole Problems

Second, set up equations to convert the number of moles of the reactant to the number of moles of each of the products using dimensional analysis.5.48 mol H2O 2 mol H2 = 5.48 mol H2

2 mol H2O

5.48 mol H2O 1 mol O2 = 2.74 mol O2

2 mol H2O

Mole-Mole Problems

How many moles of FeCl3 will be produced from 5 moles of Cl2?

2FeBr3 + 3Cl2 2FeCl3 + 3Br2

5 mol Cl2 2 mol FeCl3 = 3.33 mol FeCl3

3 mol Cl2

Mass-Mass Problems

You will be given the mass of one substance and are asked to find the mass of another substance involved in the same chemical reaction.

Mass of Unknown

Mass of given

Moles of given

Moles of Unknown

Mass-Mass Problems

Determine the mass of NaOH produced when 0.25g of Na reacts with water according to the following equation:

2Na + 2H2O 2NaOH + H2

0.25g Na 1 mol Na 2 mol NaOH 39.99711g NaOH = 22.98977gNa 2 mol Na 1 mol NaOH

0.43g NaOH

Mass-Mass Problems

How many grams of aluminum chloride will be produced from 92.00g of Cl2?

2AlBr3 + 3Cl2 3Br2 + 2AlCl3

92.00g Cl2 1 mol Cl2 2 mol AlCl3 133.341g AlCl3 =

70.906g Cl2 3 mol Cl2 1 mol AlCl3

115.3g AlCl3

Mass-Volume Problems

Used to calculate the volume of gas produced when the mass of another substance in the reaction is known.

Volume of Gaseous

UnknownMass of given

Moles of given

Moles of Unknown


How many liters of oxygen are necessary for the combustion of 340.00g of ethanol (C2H5OH) assuming that the reaction occurs at STP?

C2H5OH + 3O2 2CO2 + 3H2O

340.00g C2H5OH 1 mol C2H5OH 3 mol O2 22.4 L O2 =

46.069g C2H5OH 1 mol C2H5OH 1 mol O2

495.95 L O2


How much hydrogen gas, at STP, if formed from the reaction of 3.50g of zinc with an excess amount of hydrochloric acid?

Zn + 2HCl ZnCl2 + H2

3.50 g Zn 1 mol Zn 1 mol H2 22.4 L H2 =

65.39g Zn 1 mol Zn 1 mol H2

1.20 L H2

Volume-Volume Problems

The coefficients in a chemical equation represent the ratio of the volumes of gases involved in the reaction.

Very similar to Mole-Mole problems.

Volume-Volume Problems

What volumes of hydrogen and nitrogen gases are necessary to produce 16.0 L of ammonia gas?

N2 + 3H2 2NH3

16.0 L NH3 1 mol NH3 1 mol N2 22.4 L N2 = 8.00 L N2

22.4 L NH3 2 mol NH3 1 mol N2

16.0 L NH3 1 mol NH3 3 mol H2 22.4 L H2 = 24.0 L H2

22.4 L NH3 2 mol NH3 1 mol H2

Limiting Reactants

The amount of product that can be produced from a chemical reaction depends on the amounts of reactants available.

Stoichiometric proportion – when the quantities of reactants are available in the exact ratio described by the balanced equation.

Limiting Reactants

There is usually more of one reactant than can be used… this reaction is said to be in nonstoichiometric proportions.

The reactant that gets used up first and therefore limits the amount of product formed is called the limiting reactant.

Limiting Reactants

To solve a limiting reactant problem, you must complete two separate mass-mass problems. First calculate the mass of product formed using the given mass of one reactant. Then calculate the mass of product formed using the given mass of the other reactant.

The limiting reactant is the one that produces the smaller amount of product.

Limiting Reactant Problems

Identify the limiting reactant when 10.0 g H2O reacts with 3.5g Na to produce NaOH and H2.

2Na + 2H2O 2NaOH + H2 10.0 g H2O 1 mol H2O 2 mol NaOH 39.9971 g NaOH =

18.0153g H2O 2 mol H2O 1 mol NaOH

22.2 g NaOH


Identify the limiting reactant when 10.0 g H2O reacts with 3.5g Na to produce NaOH and H2.

2Na + 2H2O 2NaOH + H2

3.5g Na 1 mol Na 2 mol NaOH 39.9971 g NaOH =

22.98977g Na 2 mol Na 1 mol NaOH

6.1 g NaOH


NaCl reacts with Pb(NO3)2 to produce PbCl2 and NaNO3. What is the limiting reactant when 45.3g of NaCl reacts with 60.8g of Pb(NO3)2?

Pb(NO3)2 + 2NaCl PbCl2 + 2NaNO3

60.8g Pb(NO3)2 1 mol Pb(NO3)2 1 mol PbCl2 278.1054g PbCl2 =

331.2098g Pb(NO3)2 1mol Pb(NO3)2 1 mol PbCl2

51g PbCl2


NaCl reacts with Pb(NO3)2 to produce PbCl2 and NaNO3. What is the limiting reactant when 45.3g of NaCl reacts with 60.8g of Pb(NO3)2?

Pb(NO3)2 + 2NaCl PbCl2 + 2NaNO3

45.3g NaCl 1 mol NaCl 1 mol PbCl2 278.1054g PbCl2 =

58.44247g NaCl 2 mol NaCl 1 mol PbCl2

108g PbCl2


Identify the limiting rectant when 12.5 L H2S at STP is bubbled through a solution containing 24.0g KOH to form K2S and H2O.

H2S + 2KOH K2S + 2H2O12.5 L H2S 1 mol H2S 2 mol H2O 18.01528g H2O =20.1g H2O

22.4L H2S 1 mol H2S 1 mol H2O

24.0g KOH 1 mol KOH 2mol H2O 18.01528g H2O = 7.71gH2O

56.10564g KOH 2mol KOH 1 mole H2O

Percent Yield

The percent of the expected yield which was actually obtained experimentally.

Percent Yield = actual yield x 100 expected yield

Actual Yield- the amount of a product that is really obtained from a chemical reaction.

Expected Yield- the amount of product that should be produced based on calculations.

Percent Yield

Determine the percent yield for the reaction between 2.80g Al(NO3)3 and excess NaOH if 0.966g of Al(OH)3 is recovered.

Al(NO3)3 + 3NaOH Al(OH)3 + 3NaNO3

2.80gAl(NO3)3 1 mol Al(NO3)3 1mol Al(OH)3 78.00356gAl(OH)3 =

212.996gAl(NO3)3 1mol Al(NO3)3 1mol Al(OH)3

1.03g Al(OH)3

Percent Yield

Actual Yield = 0.966g Al(OH)3

Theoretical Yield = 1.03g Al(OH)3

% Yield = Actual Yield x 100

Theoretical Yield

% Yield = 0.966g x 100 = 93.8%

1.03g

unit 9 stoichiometry

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