Solution

CuSO₄–5H₂O (s) è 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) è 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) è 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) 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 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) 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?

                                  C_XH_y + [x+(y/4)] O₂ è xCO₂ + (y/2) H₂O (l)

 Solution

By comparing coefficients of C_xH_y and CO₂,
2:6 = 1:x
Therefore, x=3

By comparing coefficients of O₂ and C_xH_y,
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.

                                  C_XH_y + [x+(y/4)] O₂ è xCO₂ + (y/2) H₂O (l)

Solution

Total initial volume – Total final volume = 25cm³
(10 + V_i) – (V_f + 30) = 25cm³
(V_i – V_f) = 45cm³
Mole ratio is 1:4.5:3
By comparing coefficients of C_xH_y and CO₂,
1:3 = 1:x
Therefore, x=3

By comparing coefficients of O₂ and C_xH_y,
x+ (y/4) : 1 = 4.5:1

Because x = 3,
Therefore y =6
Formula of hydrocarbon is C₃H₆