Consider the reaction for the recovery of metallic iron: 
Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g)
Calculate the percent yield of the reaction of 6.930 g of Fe₂O₃ in excess H₂ if 2.930 g of Fe (s) was produced. The molar mass of Fe₂O₃ is 159.7 g/mol.