1. Chapter 9 Stoichiometry Test REVIEW SHEET

2. 2. RELATIVE NUMBER OF MOLES
1. A balanced chemical equation allows one to determine the _____ ________ between all compounds in the equation.
2. The coefficients in a chemical equation represent the ________ ____ ________ of reactants and products.
3. Actual yield of a product in a reaction must be determined by_________________________.
4. Knowing the mole ratio of a reactant and product in a chemical reaction would allow you to determine the __________ and ____________ of a product.
1. MOLAR RATIOS
2. RELATIVE NUMBER OF MOLES
3. EXPERIMENTS
4. MOLES AND MASS

3. 5. ONLY 1 OF THE PRODUCTS 6. SIX (2:4, 2:3, 4:3, 3:4, 3:2, 4:2) 7. 4:3
5. To determine the limiting reactant in a chemical reaction involving masses of the two reactants, how many products should be used in your molar ratios?
6. How many mole ratios can be correctly obtained from the chemical equation 2Al2O3(l) → 4Al(s) + 3O2(g)?
7. In the equation 2Al2O3 → 4Al + 3O2, what is the mole ratio of aluminum to oxygen?
8. What is the name of the reactants in a chemical reaction that restricts how much product can be made?
5. ONLY 1 OF THE PRODUCTS
6. SIX (2:4, 2:3, 4:3, 3:4, 3:2, 4:2)
7. 4:3
8. LIMITING REACTANT

4. 9. If the percentage yield for a chemical reaction is 80
9. If the percentage yield for a chemical reaction is 80.0%, and the theoretical yield is 100 grams, what is the actual yield of product?
11. What is the mole ratio of H2O to H3PO4 in the following chemical equation?
P4O10 + 6H2O → 4H3PO4
80/100 X 100 = 80% GRAMS
11. 6:4 OR 3:2

5. 8mol SiO2 = 1 = 8 mol SiC x 40g = 320 g SiC 1 1mol
12. In the formation of silicon carbide represented by the chemical equation
SiO2(s) + 3C(s) → SiC(s) + 2CO(g), 8 mol of each reactant are available for the reaction.
What substance is the excess reactant?
8mol SiO2 = 1 = 8 mol SiC x 40g = 320 g SiC
mol
8mol C = 1 = 2.67 mol SiC x 40g = g SiC
mol
C IS THE LIMITING REACTANT
SiO2 IS THE EXCESS REACTANT

6. 13. For the reaction represented by the equation
SO3 + H2O → H2SO4, what is the percentage yield if g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid?
Element Molar mass
Hydrogen g/mol
Oxygen g/mol
Sulfur g/mol
g SO3 x 1mol = 6.24mol SO3 = 1 = 6.24 mol H2SO4 x 98.09
80.07g mole =
grams of H2SO4: THEORETICAL YIELD
575g x = 93.94%
612.08g

7. O2 IS THE LIMITING REACTANT
14. If the following reaction were to take place C5H12 + O2 → CO2 + H2O
a. Write the balanced chemical equation
b. What are all of the possible mole ratios?
c. Find the limiting reactant if 3 mol of C5H12 reacts with 2.3 mol of O2
14A. C5H12 + 8O2  5CO2 + 6H20
14B. 1:8, 1:5, 1:6, 8:5, 8:6, 5:6, 8:1, 5:1, 6:1, 5:8, 6:8, 6:5
14C. 3 mol C5H12 = 5 = 15 mol CO2 x 44g = 660 g CO2
1 1mol
2.3 mol O2 = 5 = mol CO2 x 44g = g CO2
mol
O2 IS THE LIMITING REACTANT

8. 14. d. How much excess reactant would there be?
14. e. How many grams of water would you expect to form?
14D.2.3 mol O2 = 1 = mol C5H12 ACTUALLY USED UP 8
3 mol C5H mol C5H12 = excess
14E.2.3 mol O2 = 6 = mol H2O x 18g = g H2O
mol

9. Fe3N2 IS THE LIMITING REACTANT
15. If the following reaction were to take place Al + Fe3N2 → AlN + Fe
a. Write the balanced chemical equation
b. What are all of the possible mole ratios?
c. Find the limiting reactant if 1.7 mol of Al reacts with .6 mol of Fe3N2
15A. 2Al + Fe3N2  2AlN + 3Fe
15B. 2:1, 2:2, 2:3, 1:2, 1:3, 2:3, 1:2, 2:2, 3:2, 2:1, 3:1, 3:2
15C. 1.7mol Al = 3 = 2.55mol Fe x 54g = g Fe
mol
.6mol Fe3N2 = 3 = 1.8mol Fe x g = 97.2 g Fe
mol
Fe3N2 IS THE LIMITING REACTANT

10. .6mol Fe3N2 = 2 = 1.2mol Al ACTUALLY USED UP 1
15d. How much excess reactant would there be?
15e. How many grams of iron would you expect to form?
.6mol Fe3N2 = 2 = 1.2mol Al ACTUALLY USED UP 1
1.7 mol – 1.2 mol = .5mol excess Al
.6mol Fe3N2 = 3 = 1.8mol Fe x 54g = 97.2 g Fe
mol

11. 16. Calculate the percentage yield for the reaction represented by the equation
CH4 + 2O2 →2H2O + CO2
when g of CH4 react with excess O2 to produce g of CO2.
1000g CH4 x 1mol = 62.5mol CH4 = 1 = 62.5mol CO2
16g
62.5 mol CO2 x 44g = 2750g C : THEORETICAL YIELD
1 mol
2300g x = %
2750g