When 12.71 g of copper react with 8.000 g of sulfur, 14.72 g of copper (I) sulphide is produced. What is the percentage yield in this reaction?

2Cu (s) + S (g) → Cu2S (s)

Show your work.

2Cu + S = Cu₂S

n (Cu) = m (Cu) / M (Cu)

n (Cu) = 12.71/63.55=0.2 mol

n (S) = m (s) / M (S)

n (S) = 8.000/32.01=0.25 mol

Cu:S 2:1

0.2 mol : 0.25 mol copper deficiency, sulphur in excess

theoretical mass of copper (I) sulfide

m (Cu₂S) = M (Cu₂S) n (Cu) / 2

the percentage yield in the reaction is

w=m' (Cu₂S) / m (Cu₂S) = 2m' (Cu₂S) / M (Cu₂S) n (Cu)

w=2*14.72 / (159.16*0.2) = 0.925 (92.5%)