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%)