Blister copper is produced by reaction of copper oxide with copper sulphide.
2Cu₂O + Cu₂S → 6Cu + SO₂
When 2.86 × 10³ g of Cu₂O and 4.77 × 10³ g of Cu₂S are used for reaction, the mass of copper produced is _____________ g. (nearest integer)
(Atomic mass of Cu = 63.5 a.m. u, S = 32.0 a.m. u, O = 16.0 a.m. u)

Solution

Chemistry ma’am · 4 steps

Turn both masses into moles, then compare each with its coefficient to find which one runs out first.

1. Molar masses. Cu₂O: 2 × 63.5 + 16 = 143 g mol⁻¹. Cu₂S: 2 × 63.5 + 32 = 159 g mol⁻¹.
2. Moles taken. Cu₂O: 2860/143 = 20 mol. Cu₂S: 4770/159 = 30 mol.
3. Limiting reagent. The equation needs 2 mol Cu₂O for every 1 mol Cu₂S. So 20 mol Cu₂O needs only 10 mol Cu₂S, and 30 mol is available. Cu₂O runs out first, so it is the limiting reagent.

   n_(Cu) = 20 × 6/2 = 60 mol

4. So. Mass of copper = 60 × 63.5 = 3810 g.

The common slip: 7620. Taking 6 mol Cu per mol of Cu₂O instead of 6 per 2 mol gives 120 mol Cu and 7620 g, double the answer.

The idea to keep. To find the limiting reagent, divide each reactant's moles by its coefficient. The smallest value runs out first, and all product amounts come from that reactant.

Answer from the NTA final answer key. Solution by PrepWiser.