7.62g of iron reacts with 8.67g of sulfur what is the percentage yield if only 8g of Fes is produced

Fe + S = FeS (ratio is 1:1)

Molar mass of FeS = 87.9107 g/mol

(0.1364 mol Fe) x (1/1) x (87.9107 g FeS/mol) = 12.0 g FeS

Theoretical yield = 12.0 g FeS

If the actual yield of FeS = 8g

Then percent yield (8 g / 12.0 g) = 0.67 = 67%