Question

If 12.8 grams of iron (III) oxide react with 11.5 grams of carbon monoxide to produce 6.10 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 yields Fe + CO2

Answer 1

First, balance the equation

 Fe2O3 + CO ---> Fe + CO2

 Fe2O3 + 3CO yields 2Fe + 3CO2 (Balanced)

Find the limiting reactant. The ratio of iron (iii) oxide to Carbon Monoxide should be 1:3.

You can only use 1/3 as much Fe2O3 as you can CO, as it is the limiting reactant and must follow the ratio.

11.5 g / 3 = 3.833333 g Fe2O3

The ratio of Fe2O3 the reactant to pure iron the product is 1:2, so just multiply by 2

3.833333 g Fe2O3 x 2 = 7.67 g Fe

This calculated amount is the theoretical yield, but in real life, there was some spillage. Only 6.10 g was produced.

6.10 g (actual yield) / 7.67 g (theoretical yield) = .7967 x 100 = 80.0 % yield

Hope I helped!