Limiting Reactions

Question

3. In the production of copper from ore containing copper (II) sulfide, the ore is first roasted to change it to the oxide according to the following equation:

2CuS + 3O₂ ---> 2CuO + 2SO₂

_{a. If 100 g of CuS and 56 g of O2 are available, which reactant is limiting?}

_{b. What mass of CuO can be formed from the reaction of 18.7 g of CuS and 12.0 g of O2?}

(Please show work if possible so I understand how you got your answer.)

Answer 1

a) Divide by molar mass: 100g CuS * (1 mol CuS / 95.6g CuS) = 1.05 mole CuS, then modify by mole ratio from the balanced reaction: (1.05 mol CuS) * (2 mol CuO/2 mol CuS) = 1.05 mole of CuO product.

With 56g O₂ * (1 mol O₂ / 32.0g O₂) = 1.17 mole O₂ * (2 mol CuO/3 mol O₂) -> 1.17 mole of CuO. Therefore CuS is the limiting reactant, because there will be O₂ left over when all 110g of CuS are consumed.

b) 18.7g CuS * (1 mol CuS / 95.6g CuS) * (2 mol CuO/2 mol CuS) = 0.1956 mole CuO.

12.0g O₂ * (1 mol O₂ / 32.0g O₂) * (2 mol CuO/3 mol O₂) = 0.250 mole CuO. Again CuS is limiting. Finally 0.1956 mol CuO * (79.6g CuO /1 mol CuO) = 15.6g CuO.