Question

Iron (III) oxide (s) is heated in excess carbon monoxide (g) to form iron (s) and carbon dioxide (g). If the percent yield of iron was 83.0% and the actual yield was 9.87 g, how much iron (III) oxide (s) was used

Answer 1

17.1 grams Fe2O3

Step-by-step explanation:

We need to balance the chemical reaction.

I find:

1Fe2O3 + 3CO = 3CO2 + 2Fe

A little tricky, but count the individual atoms and you'll find that all are accounted for.

We are told that 9.87 grams of Fe was obtained, and that was an 83% yield. In other words, the actual amount, 9.87 grams, when divided by the theoretical amount, x, is 83.0% or 0.83:

9.87g/x = 0.83

x = 11.89 grams (the theoretical amount we should have obtained, if everything went perfectly.)

11.89 grams of Fe, at 55.85 g/mole, is (11.89 g)/(55.85 g/mole) = 0.213 moles Fe. The balanced equation says it takes 1 mole of Fe2O3 to make 2 moles of Fe, a ratio of (1 mole Fe2O3/2 moles Fe)

(0.213 moles Fe)*(1 mole Fe2O3/2 moles Fe) = 0.107 moles Fe2O3

(0.107 moles Fe2O3)*(159.7 grams Fe2O3/mole Fe2O3) = 17.1 grams Fe2O3.