Sample Problem 3.10

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
4
3
1.00x102g N2H4
3.12mol N2H4
N2H4 is the limiting reactant because it produces
less product, N2, than does N2O4.
3.12mol N2H4
4.68mol N2
4.68mol N2
131g N2
2.00x102g N2O4
2.17mol N2O4
2.17mol N2O4
6.51mol N2
3
The effect of side reactions on yield.
Figure 3.9
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 100.0 kg of sand is processed, 51.4
kg of SiC is recovered. What is the percent
yield of SiC from this process?
PLAN
SOLUTION
write balanced equation
100.0 kg SiO2
1664 mol SiO2
find mol reactant product
mol SiO2 mol SiC 1664
find g product predicted
1664 mol SiC
66.73 kg
actual yield/theoretical yield x 100
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
Methanol burns in air according to the equation
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
molar mass CH3OH
molar mass H2O
coefficients chemical equation
209 g CH3OH
235 g H2O
7
Limiting Reagent
Reactant used up first in the reaction.
NO is the limiting reagent
O2 is the excess reagent
8
In one process, 124 g of Al are reacted with 601
g of Fe2O3
Calculate the mass of Al2O3 formed.
OR
367 g Fe2O3
124 g Al
Have more Fe2O3 (601 g) so Al is limiting reagent
9
Use limiting reagent (Al) to calculate amount of
product that can be formed.
234 g Al2O3
124 g Al
At this point, all the Al is consumed and Fe2O3
remains in excess.
10
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.