An ore contains 12 % by mass of metal X oxide, X₂O₃ (Mₓ = 54 g mol⁻¹). A 500 g sample of the ore is reduced in two separate processes.

Process 1: X₂O₃ + 3 C → 2 X + 3 CO, using excess carbon.

Process 2: X₂O₃ + 2 Al → 2 X + Al₂O₃, using 30 g of aluminium (Mₐₗ = 27 g mol⁻¹).

(a) Calculate the theoretical mass of metal X obtainable from the ore.

(b) If the actual mass of X obtained from Process 1 is 38 g, determine the percentage yield.

(c) If Process 2 gives an actual mass of 40 g of X, determine its percentage yield and state which process is more efficient.

Question

An ore contains 12 % by mass of metal X oxide, X₂O₃ (Mₓ = 54 g mol⁻¹). A 500 g sample of the ore is reduced in two separate processes.

Process 1: X₂O₃ + 3 C → 2 X + 3 CO, using excess carbon.

Process 2: X₂O₃ + 2 Al → 2 X + Al₂O₃, using 30 g of aluminium (Mₐₗ = 27 g mol⁻¹).

(a) Calculate the theoretical mass of metal X obtainable from the ore.

(b) If the actual mass of X obtained from Process 1 is 38 g, determine the percentage yield.

(c) If Process 2 gives an actual mass of 40 g of X, determine its percentage yield and state which process is more efficient.

Accepted Answer

Theoretical mass = 41.5 g; Yield Process 1 = 91.6 %; Yield Process 2 = 96.4 %; Process 2 is more efficient.

Igcse Model Questions Vol1 2020 Chemistry Question 31