1. Sample Problem 3.10
Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem
PROBLEM:
A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed?
PLAN:
We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given.
In this case one of the reactants is in molar excess and the other will limit the extent of the reaction.
mass of N2H4
mass of N2O4
limiting mol N2
divide by M
multiply by M
mol of N2H4
mol of N2O4
g N2
molar ratio
mol of N2
mol of N2

2. Sample Problem 3.10
Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem
continued
SOLUTION: 2
N2H4(l) + N2O4(l) N2(g) H2O(l) 3 4
mol N2H4
32.05g N2H4
1.00x102g N2H4
= 3.12mol N2H4
N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.
3 mol N2
2mol N2H4
3.12mol N2H4
= 4.68mol N2
mol N2
28.02g N2
4.68mol N2
= 131g N2
mol N2O4
92.02g N2O4
2.00x102g N2O4
= 2.17mol N2O4
3 mol N2
mol N2O4
2.17mol N2O4
= 6.51mol N2

3. C A + B D The effect of side reactions on yield. Figure 3.9
(main product)
A + B
(reactants) D
(side products)

4. Sample Problem 3.11
Calculating Percent Yield
PROBLEM:
Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process?
PLAN:
SOLUTION:
SiO2(s) + 3C(s) SiC(s) + 2CO(g)
write balanced equation
103 g SiO2
kg SiO2
mol SiO2
60.09 g SiO2
100.0 kg SiO2
= 1664 mol SiO2
find mol reactant & product
mol SiO2 = mol SiC = 1664
find g product predicted
40.10 g SiC
mol SiC kg
103g
1664 mol SiC
= kg
actual yield/theoretical yield x 100
51.4 kg
66.73 kg
percent yield
x 100
=77.0%

5. Amounts of Reactants and Products
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units

6. Limiting Reagent: Reactant used up first in the reaction.
2NO + O NO2
NO is the limiting reagent
O2 is the excess reagent

7. Have more Fe2O3 (601 g) so Al is limiting reagent
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2 Al + Fe2O Al2O3 + 2 Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent

8. 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
At this point, all the Al is consumed and Fe2O3 remains in excess.

9. Reaction Yield Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100%