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### 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) è CuSO(s) + 5H2O (s)

Let the mass of CuSO45H2O be x g
Number of moles of CuSO= Number of moles of CuSO45H2
= x/249.6

MgSO47H2O (s) è MgSO(s) + 5H2O (s)
Mass of MgSO47H2O is (5-x) g
Number of moles of MgSO= Number of moles of MgSO47H2O
= (5-x)/246.4

Total mass of solid after heating = 3.00g
Mass of CuSO+ Mass of MgSO=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) è 2Al(OH)3 (s) + 3H2S (g)

Number of moles of Al2S= 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, Al2Sis 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) + 3H(g) è 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 10cmof 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

CH(s) + 2O(g) è CO2 (g) + 2H2O (l)
Stoichiometric Coefficient is 1:2:1:2

Volume of CO2 produced is 10cm3
2C2H6 (g) + 7O2 (g) è 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 è xCO2 + (y/2) H2O (l)

 Initial Volume 20cm3 100cm3 0cm3 - Final Volume 0cm3 10cm3 (70-10)cm3 - Reacting Volume 20cm3 90cm3 60cm3 - Mole Ratio 2 9 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 è xCO2 + (y/2) H2O (l)

 Initial Volume 10cm3 Vicm3 0cm3 - Final Volume 0cm3 Vfcm3 30cm3 - Reacting Volume 20cm3 (Vi –Vf ) cm3 60cm3 - Mole Ratio 1 ? 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