Chemical Calculations Challenging
Questions
Credits: Thanks to Nicholas for contributing the questions!
Q1. A mixture of CuSO4– 5H2O and MgSO4– 7H2O is heated until a mixture of the
anhydrous salt is obtained. If 5.00g of the mixture gives 3.00g of the anhydrous salts, what is
the percentage by mass of CuSO4– 5H2O in the mixture?
Solution
CuSO4– 5H2O (s)  CuSO4 (s) + 5H2O (s)
Let the mass of CuSO4– 5H2O be x g
Number of moles of CuSO4 = Number of moles of CuSO4– 5H2O
= x/249.6
MgSO4– 7H2O (s)  MgSO4 (s) + 5H2O (s)
Mass of MgSO4– 7H2O is (5-x) g
Number of moles of MgSO4 = Number of moles of MgSO4– 7H2O
= (5-x)/246.4
Total mass of solid after heating = 3.00g
Mass of CuSO4 + Mass of MgSO4 =3.00g
[ (x/249.6) x 159.6 + ((5-x)/246.4) x 120.4 ] = 3
Solving for x,
x = 3.69
Percentage of CuSO4– 5H2O = 3.69/5 x100
= 73.8%
Q2. Solid aluminium sulfide reacts with water to give aluminium hydroxide and
hydrogen sulfide gas. Write a balanced equation for this reaction. What is the
maximum mass of H2S that can form when 158g of aluminium sulfide reacts with
131g of water? Which is the limiting reactant? Calculate the number of moles of
excess reagent remaining at the end of the reaction.
Solution
Al2S3 (s) + 6H2O (l)  2Al(OH)3 (s) + 3H2S (g)
Number of moles of Al2S3 = 158/150.3
= 1.051mol
Number of moles of H2O = 131/18
= 7.28mol
Stoichiometric Coefficient is 1:6:2:3
1.05 x 6 = 6.306
Therefore, Al2S3 is the limiting reactant.
7.28 – 6.306 = 0.974mol
Q3. What is the percentage yield of a reaction in which 41.5g of Tungsten(VI) Oxide
reacts with excess hydrogen gas to produce metallic tungsten and 9.50ml of water
(density = 1.00gml-1)
Solution
WO3 (s) + 3H2 (g)  W (s) + 3H2O (l)
WO3 is the limiting reactant.
Number of moles of WO3 = 41.5/232
= 0.1789mol
Theoretical mass of W formed = 0.1789 x 184
= 32.92g
Actual number of moles of W formed = 1/3 x 9.50/18
= 0.1759mol
Actual mass of W formed = 0.1759 x 184
= 32.37g
Percentage yield of W = 32.37/32.97 x 100%
= 98.3%
Q4. A mixture of 10cm3 of methane and 10cm3 of ethane was sparked with an
excess of oxygen. After cooling to room temperature, the residual gas was passed
through aqueous potassium hydroxide. What volume of gas was absorbed by the
alkali?
Solution
CH4 (s) + 2O2 (g)  CO2 (g) + 2H2O (l)
Stoichiometric Coefficient is 1:2:1:2
Volume of CO2 produced is 10cm3
2C2H6 (g) + 7O2 (g)  4CO2 (g) + 6H2O (l)
Stoichiometric Coefficient is 2:7:4:6
Volume of CO2 produced is 20cm3
Total volume of CO2 produced is 30cm3
Q5. 20cm3 of a gaseous hydrocarbon was mixed with 100cm3 if oxygen so that the
hydrocarbon was completely burnt. The volume of gas remaining at the end of
combustion was 70cm3. After passing over potassium hydroxide, this volume was
reduced to 10cm3. All gases were measured at 25ºC and at the same pressure.
What is the formula of the hydrocarbon?
CXHy + [x+(y/4)] O2  xCO2 + (y/2) H2O (l)
Initial Volume
Final Volume
Reacting Volume
Mole Ratio
20cm3
0cm3
20cm3
2
100cm3
10cm3
90cm3
9
0cm3
(70-10)cm3
60cm3
6
-
Solution
By comparing coefficients of CxHy and CO2,
2:6 = 1:x
Therefore, x=3
By comparing coefficients of O2 and CxHy,
x+ (y/4) : 1 = 9:2
Because x = 3,
Therefore y =6
Formula of hydrocarbon is C3H6
Q6. 10cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen. A
contraction of 25cm3 in volume occurs after the combustion. On passing the
gaseous products through aqueous sodium hydroxide, a further contraction of
30cm3 occurs. Deduce the formula of the hydrocarbon. (All volumes were
measured at r.t.p.
CXHy + [x+(y/4)] O2  xCO2 + (y/2) H2O (l)
Initial Volume
Final Volume
Reacting Volume
Mole Ratio
10cm3
0cm3
20cm3
1
Vicm3
Vfcm3
(Vi –Vf ) cm3
?
0cm3
30cm3
60cm3
3
-
Solution
Total initial volume – Total final volume = 25cm3
(10 + Vi) – (Vf + 30) = 25cm3
(Vi – Vf) = 45cm3
Mole ratio is 1:4.5:3
By comparing coefficients of CxHy and CO2,
1:3 = 1:x
Therefore, x=3
By comparing coefficients of O2 and CxHy,
x+ (y/4) : 1 = 4.5:1
Because x = 3,
Therefore y =6
Formula of hydrocarbon is C3H6