Worked Solutions
Chapter 16
Question 63
Ethanal was prepared in the school laboratory by the oxidation of ethanol using acidified
sodium dichromate(VI), Na2Cr2O7.2H2O, according to the equation:
3C2H5OH + Cr2O72- + 8H+  3CH3CHO + 2Cr3+ + 7H2O
8.84 g of sodium dichromate(VI), and 6.9 cm3 of ethanol were used, and 1.6 g of ethanal were
formed. The density of ethanol is 0.8 g cm-3.
(a) Show that sodium dichromate(VI) is the limiting reactant.
(b) Calculate the percentage yield of ethanal.
Answer:
(a) Moles of sodium dichromate present initially = 8.84 / 298 = 0.0297
Mass of ethanol present initially = density X volume = 6.9 X 0.8 g
= 5.52 g
Moles of ethanol present initially = 5.52 / 46 = 0.12
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
3 moles
1 mole
0.12 moles 0.04 moles
3 moles
0.12 moles
2 moles
7 moles
0.08 moles 0.28 moles
Since the 0.12 moles of ethanol that are present initially would react fully with 0.04 moles of
sodium dichromate, and there are only 0.0297moles of sodium dichromate present initially, it
is clear that it is the ethanol that is present in excess.
Sodium dichromate is the limiting reactant.
(b) 3C2H5OH(l)
+ Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l)
3 X 0.0297 moles 0.0297 moles
3 X 0.0297 moles
Theoretical yield of ethanal = 0.0891 moles = 0.0891 X 44 g = 3.92 g
Actual yield of ethanal = 1.6 g
Actual yield X 100
1.6 X 100
Percentage yield =  % =  %
Theoretical yield
3.92
= 40.82%
Question 64
Ethanoic acid was prepared in the school laboratory by the oxidation of ethanol using
acidified sodium dichromate(VI), Na2Cr2O7.2H2O, according to the equation:
3C2H5OH + 2Cr2O72- + 16H+  3CH3COOH + 4Cr3+ + 11H2O
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8.84 g of sodium dichromate(VI), and 2.3 cm3 of ethanol were used, and 1.7 g of ethanoic
acid were formed. The density of ethanol is 0.8 g cm-3.
(a) Show that ethanol is the limiting reactant.
(b) Calculate the percentage yield of ethanoic acid.
Answer:
(a) Moles of sodium dichromate present initially = 8.84 / 298 = 0.0297
Mass of ethanol present initially = density X volume = 2.3 X 0.8 g
= 1.84 g
Moles of ethanol present initially = 1.84 / 46 = 0.04
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
3 moles
2 moles
0.04 moles 0.0267 moles
3 moles
0.04 moles
Since the 0.04 moles of ethanol that are present initially would react fully with 0.0267 moles
of sodium dichromate, and there are 0.0297moles of sodium dichromate present initially, it is
clear that it is the sodium dichromate that is present in excess.
Ethanol is the limiting reactant.
(b) 3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
0.04 moles 0.0267 moles
0.04 moles
Theoretical yield of ethanoic acid = 0.04 moles = 0.04 X 60 g = 2.4 g
Actual yield of ethanoic acid = 1.7 g
Actual yield X 100
1.7 X 100
Percentage yield =  % =  %
Theoretical yield
2.4
= 70.83%
Question 65
Ethanal was prepared by a group of students by the oxidation of ethanol using acidified
sodium dichromate(VI), Na2Cr2O7.2H2O. The balanced equation is:
3C2H5OH + Cr2O72- + 8H+  3CH3CHO + 2Cr3+ + 7H2O
12.5 g of sodium dichromate(VI), and 7.73 g of ethanol were used, and 1.95 g of ethanal were
formed.
(a) Show clearly that ethanol is in excess in the experiment.
(b) Calculate the percentage yield of ethanal.
Answer:
(a) Moles of sodium dichromate present initially = 12.5 / 298 = 0.042
Mass of ethanol present initially = 7.73 g
Moles of ethanol present initially = 7.73 / 46 = 0.168
2
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
3 moles
1 mole
3 moles
0.168 moles 0.056 moles
2 moles
7 moles
0.168 moles
Since the 0.168 moles of ethanol that are present initially would react fully with 0.056 moles
of sodium dichromate, and there are only 0.042 moles of sodium dichromate present initially,
it is clear that it is the ethanol that is present in excess.
+ Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l)
(b) 3C2H5OH(l)
3 X 0.042 moles
3 X 0.042 moles
0.042 moles
Theoretical yield of ethanal = 0.126 moles = 0.126 X 44 g = 5.544 g
Actual yield of ethanal = 1.95 g
Actual yield X 100
1.95 X 100
Percentage yield =  % =  %
Theoretical yield
5.544
= 35.17%
Question 66
When 6.2 g of sodium dichromate(VI), Na2Cr2O7.2H2O were reacted with 1.29 g of ethanol,
in the presence of sulfuric acid, 1.2 g of ethanoic acid were formed. The equation for the
reaction is: 3C2H5OH + 2Cr2O72- + 16H+  3CH3COOH + 4Cr3+ + 11H2O
(a) Show clearly that sodium dichromate(VI) is present in excess in the experiment.
(b) Calculate the percentage yield of ethanoic acid.
Answer:
(a) Moles of sodium dichromate present initially = 6.2 / 298 = 0.021
Mass of ethanol present initially = 1.29 g
Moles of ethanol present initially = 1.29 / 46 = 0.028
3C2H5OH(l)
3 moles
+ 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
2 moles
0.028 moles 0.0187 moles
3 moles
0.028 moles
Since the 0.028 moles of ethanol that are present initially would react fully with 0.0187 moles
of sodium dichromate, and there are 0.021moles of sodium dichromate present initially, it is
clear that it is the sodium dichromate that is present in excess.
(b) 3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
3 moles
2 moles
0.028 moles 0.0187 moles
3 moles
0.028 moles
Theoretical yield of ethanoic acid = 0.028 moles = 0.028 X 60 g = 1.68 g
Actual yield of ethanoic acid = 1.2 g
3
Actual yield
1.2 X 100
Percentage yield =  % =  %
Theoretical yield
1.68
= 71.43%
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