Question

If 8.65 grams of iron (III) oxide reacts with 7.85 grams of carbon monoxide to produce 5.25 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem.

unbalanced equation: Fe2O3 + CO Fe + CO2

Answer 1

balance the equation:
Fe2O3 + 3 CO -> 2 Fe + 3 CO2
weight of iron(III) oxide * molar mass of oxide * mole ratio of oxide to pure iron * molar mass of pure iron = theoretical yield
(8.65g Fe2O3)*(1 mol FE2O3/159.6882g FE2O3) *(2 mol Fe/1 mol Fe2O3)*(53.845g Fe/ a mol Fe) =6.05g Fe
The carbon monoxide will not be a problem because the iron(III) oxide is the limiting reactant; work if you need it:
(8.65g Fe2O3)*(1 mol FE2O3/159.6882g FE2O3) = 0.0542mol Fe2O3
(7.85g CO)*(1 mol CO/28.0104g CO) = 0.280 mol CO
The percent yield we now find by dividing the actual yield (the yield given) by the theoretical yield:
(5.25g Fe)/(6.05g Fe) x 100 = 86.8% yield

Answer 2

thanks man alot of help