If 12.6 grams of iron (III) oxide reacts with 9.65 grams of carbon monoxide to produce 7.23 g of pure iron, what are the theoret
ical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem. unbalanced equation: Fe2O3 + CO ñyieldsî/ Fe + CO2
The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂. Next step is to convert everything to moles. 12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃ 9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO The third step is to determine the limiting and excess reactants. 0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO Therefore Fe₂O₃ is the limiting reagent while CO is in excess.
0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s) 0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield %yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
