Unit 2 – Chapter 5 – Quantities Review
Converting from moles of one compound to moles of another
 Write a balanced chemical equation for the reaction
 Use molar ratios from the balanced chemical equation
Ex. Given the following chemical equation
N2(g)
+
3H2(g)

2NH3(g)
If 4.50 mol of N2 reacts with an excess of H2, how many moles of NH3 are
produced?
4.50 mol N2 x 2 mol NH3 = 9.00 mol of NH3
1 mol N2
Therefore, 9.50 mol of NH3 are produced from 4.50 mol of N2.
Practice Problems:
1. Nitromethane, CH3NO2 (l), is a fuel commonly burned in drag racing
vehicles. It combusts in oxygen to produce carbon dioxide gas, water
vapour, and nitrogen dioxide gas. Write a balanced chemical equation.
Calculate the number of moles of water produced if 1.50 moles of
methane are burned in excess oxygen.
2. Aluminum metal reacts in copper(II) chloride solution to produce aluminum
chloride solution and copper metal. Write a balanced chemical equation.
Calculate the number of moles of copper produced if 2.20 moles of
aluminum reacts with excess copper(II) chloride.
Mass – Mass Calculations



Write a balanced chemical equation
Must use molar ratios to convert from one compound to another
Therefore must convert from mass  moles  moles  mass
Mass of
compound
A
Moles of
compound
A
Molar Ratio
Moles of
compound
B
Mass of
compound
B
Practice Problems:
3. Bauxite ore contains aluminum oxide, which is decomposed using
electricity to produce aluminum metal and oxygen. What mass of
aluminum metal can be produced from 125 g of aluminum oxide?
4. Determine the mass of oxygen required to completely burn 10.0 g of
propane, C3H8(g).
5. Calculate the mass of lead(II) chloride precipitate produced when 2.57 g of
sodium chloride in solution reacts in a double displacement reaction with
excess aqueous lead(II) nitrate.
6. Predict the mass of hydrogen gas produced when 2.73 g of aluminum
reacts in a single displacement reaction with excess sulfuric acid.
Determining the Limiting Reagent:




Assume one of the reactants is the limiting reagent (Reactant A)
Use moles of Reactant A to calculate moles of Reactant B required (molar
ratio)
If moles of Reactant B calculated is less than moles of Reactant B
available (from question) then Reactant A is limiting
If moles of Reactant B calculated is greater than moles of Reactant B
available then Reactant B is limiting
Practice Problems:
7. Complete balanced equations for the following reactions and identify the
limiting and excess reagents for each of the pairs of reactants. How much
excess is present in each case?
a. Zn(s)
+
0.42 mol
b. Cl2(aq)
10 mmol
CuSO4(aq) 
0.22 mol
+
NaI(aq) 
10 mmol
c. AlCl3(aq) + NaOH(aq) 
20 g
19 g
Determining mass of products produced



Determine Limiting Reactant
Use moles of limiting reactant and molar ratio to determine moles of
product
Convert moles of product to mass of product
Practice Problems:
8. In a reaction to produce zinc sulfide, 5.00 g of zinc and 3.00 g of sulfur, S8,
are available. The balanced equation is given below:
8 Zn (s) + S8(s)  8 ZnS(s)
(a) Identify the limiting reagent.
(b) Calculate the mass of zinc sulfide formed.
9. Propane, C3H8(g), used in gas barbecues, burns in oxygen to produce carbon
dioxide and water according to the following equation:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
In a reaction, 100.7 g of propane and 367.4 g of oxygen gas are available.
(a) Identify the limiting reagent in this reaction.
(b) Calculate the mass of water formed.
Percent Yield
Percent Yield = actual yield/theoretical yield x 100%
Practice Problem:
10. In an experiment, 5.00 g of silver nitrate is added to a solution containing
an excess of sodium bromide. It was found that 5.03 g of silver bromide is
obtained.
(a) Write a balanced equation for the reaction.
(b) What is the theoretical yield of silver bromide?
(c) What is the actual yield of silver bromide in the experiment?
(d) What is the percentage yield of the experiment?
Answers:
1. 2.25 mol
2. 3.30 mol
3. 66.2 g
4. 36.3 g
5. 6.12 g
6. 0.306 g
7. (a) 0.20 mol excess Zn(s)
(b) 5 mmol excess Cl2(aq)
(c) 0.025 mol excess NaOH(aq)
8. (a) Zn(s)
(b) 7.45 g
9. (a) C3H8(g)
(b) 164.6 g
10. (b) 5.53 g
(c) 5.03 g
(d) 91.0%