Answer to Question #206163 in General Chemistry for Ali

Question

Metal oxides are reduced in reactions that produce pure metal. If the percent yield of iron is 78% when hematite (Fe2O3) is reduced in a blast furnace and 72% when magnetite (Fe3O4) is reduced which ore would produce the greatest amount of iron per metric ton (1000kg) of reactant used?

Answer 1

The balanced equation 2Fe₂O₃ ---> 4Fe + 3O₂ for hematite.

Mol. Mass Fe₂O₃ = 159.7 x 2 =319.4 kg

Atomic mass Fe = 55.845 x 4 = 223.4 kg

If all 1000 kg Fe₂O₃ reacted, the mass of Fe produced would be 223.4 kg Fe/319.4 kg Fe₂O₃ x 1000 kg Fe₂O₃ = 699.4 kg Fe. As the yield is 72%, the actual mass of Fe produced is 699.4 kg x 0.72 = 503.6 kg.

The balanced equation Fe₃O₄ ------> 3Fe + 2O₂ for magnetite

Mol. Mass Fe₃O₄ = 55.845 x 3 = 231.5 kg

Atomic mass Fe = 55.845 x 3 = 167.5 kg

If all 1000 kg Fe₃O₄ reacted, the mass of Fe produced would be 167.5 kg/231.5 kg x 1000 kg = 723.5 kg.

As the yield was 78%, the actual mass of Fe is 564.4 kg. Magnetite produces the greater mass of Fe