Question

If 0.450 moles of iron III oxide (Fe2O3) are allowed to react with an excess of aluminum (Al) and 43.6 grams of iron (Fe) is produced, what is the percent yield of iron? 2Al + Fe2O3 2Fe + Al2O3

Answer 1

0.45 moles fe2o3x 2moles fe/1 mole fe2o3 = 0.90moles fe

0.90 moles fe x 55.8g fe/1mole fe = 50.22

theoretical yield/actual yield = percent yield

43.6g/50.22g = 86.8%.