Limiting Reagents and Percentage Yield Worksheet
1.
Consider the reaction
I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g)
a) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO.
Determine the mass of iodine I2, which could be produced?
80 g I2O5
1 mol I2O5
1 mol I2
XS
1
333.8 g I2O5
1 mol I2O5
28 g CO
1
1 mol CO
28 g CO
1 mol I2
5 mol CO
253.8 g I2
1 mol I2
= 50.8 g I2
b) If, in the above situation, only 0.160 moles, of iodine, I2 was produced.
i) what mass of iodine was produced?
253.8 g I2
0.160 mol I2
= 40.6 g I2
1 mol I2
ii) what percentage yield of iodine was produc ed.
40.6 g I2
x 100 =
79.9 %
50.8 g I2
2.
Zinc and sulphur react to form zinc sulphide according to the equation.
Zn + S  ZnS
If 25.0 g of zinc and 30.0 g of sulphur are mixed,
a) Which chemical is the limiting reactant?
25.0 g Zn
1 mol Zn
1 mol ZnS
=.382 mol ZnS
1
65.41 g Zn
1 mol Zn
30.0 g S
1
1 mol S
32.1g S
1 mol ZnS
1 mol S
=.934mol ZnS
b) How many grams of ZnS will be formed?
.382 mol ZnS
97.51 g ZnS
1 mol ZnS
XS
= 37.2 g ZnS
c) How many grams of the excess reactant will remain after the reaction is over?
.934mol ZnS -.382 mol ZnS= .552mol ZnS
.552 mol ZnS
3.
1 mol S
1 mol ZnS
32.1g S
1 mol S
LR
= 17.7g S
Which element is in excess when 3.00 grams of Mg is ignited in 2.20 grams of pure oxygen?
What mass is in excess? What mass of MgO is formed?
2Mg + O2  2MgO
3.00 g Mg
1 mol Mg
2 mol MgO
40.3 g MgO
= 4.98 g MgO
1
24.3 g Mg
2 mol Mg
1 mol MgO
2.20 g O2
1
.561 g MgO
1
4.
6.
2 mol MgO
1mol O2
40.3 g MgO
1 mol MgO
= 5.54 g MgO
5.54 g MgO - 4.98 g MgO = .561 g MgO
1 mol MgO
1mol O2
32.0 g O2
= .223 g O2
40.3 g MgO
2 mol MgO
1 mol O2
How many grams of Al2S3 are formed when 5.00 grams of Al is heated with 10.0 grams S?
2Al + 3S  Al2S3
5.00 g Al
1 mol Al
1 mol Al2S3
150.17 g Al2S3
= 13.9 g Al2S3
1
26.98 g Al
2 mol Al
1 mol Al2S3
10.0 g S
1
5.
1 mol O2
32.0 g O2
1 mol S
32.07 g S
1 mol Al2S3
3 mol S
150.17 g Al2S3
1 mol Al2S3
= 15.6 g Al2S3
When MoO3 and Zn are heated together they react
3 Zn(s) + 2 MoO3(s) ----------> Mo2O3(s) + 3 ZnO(s)
What mass of ZnO is formed when 20.0 grams of MoO3 is reacted with 10.0 grams of Zn?
20.0 g MoO3
1 mol MoO3
3 mol ZnO
XS
1
143.94g MoO3
2mol MoO3
10.0 g Zn
1 mol Zn
3 mol ZnO
81.41g ZnO
12.4= g ZnO
1
65.41g Zn
3 mol Zn
1 mol ZnO
Silver nitrate, AgNO3, reacts with ferric chloride, FeCl3, to give silver chloride, AgCl, and ferric
nitrate, Fe(NO3)3. In a particular experiment, it was plannned to mix a solution containing 25.0
g of AgNO3 with another solution containing 45.0 grams of FeCl3.
a) Write the chemical equation for the reaction.
3AgNO3 + FeCl3  3AgCl + Fe(NO3)3
b) Which reactant is the limiting reactant?
25.0 g AgNO3
1 mol AgNO3
3 mol AgCl
= 0.147mol AgCl
1
169.87g AgNO3 3 mol AgNO3
45.0 g FeCl3
1
1 mol FeCl3
162.2 g FeCl3
3 mol AgCl
1 mol FeCl3
= 0.832mol AgCl
c) What is the maximum number of moles of AgCl that could be obtained (see b)?
d) What is the maximum number of grams of AgCl that could be obtained?
0.147mol AgCl
143.32g AgCl
= 21.1 g AgCl
1
1 mol AgCl
e) How many grams of the reactant in excess will remain after the reaction is over?
0.832mol AgCl - 0.147mol AgCl = .685 mol AgCl
0.685 mol AgCl
1
1 mol FeCl3
3 mol AgCl
162.2 g FeCl3
1 mol FeCl3
= 37.0 g FeCl3
7.
Solid calcium carbonate, CaCO3, is able to remove sulphur dioxide from waste gases by the
reaction:
CaCO3 + SO2 + other reactants ------> CaSO3 + other products
In a particular experiment, 255 g of CaCO3 was exposed to 135 g of SO2 in the presence of an
excess amount of the other chemicals required for the reaction.
a) What is the theoretical yield of CaSO3?
1 mol CaCO3
1 mol CaSO3
255 g CaCO3
1
100.08g CaCO3 1 mol CaCO3
1 mol SO2
1 mol CaSO3
135 g SO2
1
64.07g SO2
1 mol SO2
120.15g CaSO3
1 mol CaSO3
120.15g CaSO3
1 mol CaSO3
= 306g CaSO3
= 253g
CaSO3
b) If only 198 g of CaSO3 was isolated from the products, what was the precentage yield of
CaSO3 in this experiment?
% Y =
8.
A/t x 100
198 g CaSO3 / 253g CaSO3 x 100= 78.2%
A research supervisor told a chemist to make 100 g of chlorobenzene from the reaction of
benzene with chlorine and to expect a yield no higher that 65%. What is the minimum quantity
of benzene that can give 100 g of chlorobenzene if the yield is 65%? The equation for the
reaction is:
C6H6 + Cl2 -----------> C6H5Cl + HCl
benzene
chlorobenzene
% Y=A/t x 100 65 % = 100 g C6H5Cl/ t t = 153.8 g C6H5Cl
153.8 g C6H5Cl
1
1 mol C6H5Cl
112.45 g C6H5Cl
1 mol C6H6
1 mol C6H5Cl
78 g C6H6
1 mol C6H6
=106.7 g C6H6
9.
10.
11.
12.
13.
Certain salts of benzoic acid have been used as food additives for decades. The potassium salt of
benzoic acid, potassium benzoate, can be made by the action of potassium permanganate on
toluene.
C7H8 + 2 KMnO4 -------> KC7H5O2 + 2 MnO2 + KOH + H2O
toluene
potassium
benzoate
If the yield of potassium benzoate cannot realistically be expected to be more than 68%, what is
the minimum number of grams of toluene needed to achieve this yield while producing 10.0 g of
KC7H5O2?
% Y=A/t x 100
Aluminum dissolves in an aqueous solution of NaOH according to the following reaction:
2 NaOH + 2 Al + 2 H2O -----> 2 NaAlO2 + 3 H2
If 84.1 g of NaOH and 51.0 g of Al react:
i) Which is the limiting reagent?
ii) How much of the other reagent remains?
iii) What mass of hydrogen is produced?
Dimethylhydrazine, (CH3)2NNH2, was used as a fuel for the Apollo Lunar Descent Module, with
N2O4 being used as the oxidant. The products of the reaction are H2O, N2, and CO2.
i) Write a balanced chemical equation for the combustion reaction.
ii) If 150 kg of (CH3)2NNH2 react with 460 kg of N2O4, what is the theoretical yield of N2?
iii) If a 30 kg yield of N2 gas represents a 68% yield, what mass of N2O4 would have been used
up in the reaction?
Magnesium metal reacts quantitatively with oxygen to give magnesium oxide, MgO. If 5.00 g of
Mg and 5.00 g of O2 are allowed to react, what weight of MgO is formed, and what weight of
which reactant is left in excess?
Adipic acid, C6H10O4, is a raw material for the making of nylon and it can be prepared in the
laboratory by the following reaction between cyclohexene, C6H10, and sodium dichromate,
Na2Cr2O7 in sulphuric acid.
3 C6H10(l) + 4 Na2Cr2O7(aq) + 16 H2SO4(aq) --------->
3 C6H10O4(aq) + 4 Cr2(SO4)3(aq) + 4 Na2SO4(aq) + 16 H2O
There are side reactions. These plus losses of product during its purification reduce the overall
yield. A typical yield of purified adipic acid is 68.6%.
(a) To prepare 12.5 grams of adipic acid in 68.6% yield requires how many grams of
cyclohexene?
(b) The only available supply of sodium dichromate is its dihydrate, Na2Cr2O7.2H2O. (Since the
reaction occurs in an aqueous medium, the water in the dihydrate causes no problems, but it
does contribute to the mass of what is taken of this reactant). How many grams of this
dihydrate are also required in the preparation of 12.5 grams of adipic acid in a yield of 68.6%?