Transcript Slide 1

Sample Problem 3.10
Calculating Amounts of Reactant and Product in a LimitingReactant Problem
PROBLEM:
PLAN:
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?
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
divide by M
mol of N2H4
multiply by M
mol of N2O4
molar ratio
mol of N2
limiting mol N2
mol of N2
g N2
Sample Problem 3.10
Calculating Amounts of Reactant and Product in a LimitingReactant Problem
continued
SOLUTION:
1.00x102g N2H4
2 N2H4(l) + N2O4(l)
mol N2H4
32.05g N2H4
3.12mol N2H4
2.00x102g
N2O4
2.17mol N2O4
3 mol N2
2mol N2H4
mol N2O4
92.02g N2O4
3 mol N2
mol N2O4
3 N2(g) + 4 H2O(l)
= 3.12mol N2H4
N2H4 is the limiting reactant
because it produces less
product, N2, than does N2O4.
= 4.68mol N2
4.68mol N2
= 2.17mol N2O4
= 6.51mol N2
28.02g N2
mol N2
= 131g N2
Figure 3.9
The effect of side reactions on yield.
A +B
C
(reactants)
(main product)
D
(side products)
Sample Problem 3.11
PROBLEM:
Calculating Percent Yield
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:
write balanced equation
find mol reactant & product
SOLUTION:
SiO2(s) + 3C(s)
100.0 kg SiO2
SiC(s) + 2CO(g)
103 g SiO2
mol SiO2
kg SiO2
60.09 g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664 mol SiC
mol SiC
actual yield/theoretical yield x 100
51.4 kg
percent yield
40.10 g SiC kg
66.73 kg
103g
x 100 =77.0%
= 66.73 kg
Amounts of Reactants and Products
1.
Write balanced chemical equation
2.
Convert quantities of known substances into moles
3.
Use coefficients in balanced equation to calculate the number of moles of
the sought quantity
4.
Convert moles of sought quantity into desired units
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Limiting Reagent:
Reactant used up first in
the reaction.
2NO + O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
6
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2 Al + Fe2O3
Al2O3 + 2 Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
mol Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
Start with 124 g Al
mol Al needed
1 mol Fe2O3
2 mol Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
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Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
g Al2O3
Al2O3 + 2Fe
1 mol Al2O3
2 mol Al
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
At this point, all the Al is consumed and Fe2O3
remains in excess.
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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.
Actual Yield
% Yield =
x 100%
Theoretical Yield
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