A reaction mixture initially contains 22.02 g Fe2O3 and 14.66 g CO. Assume that the reaction will progress to 100% completion. What mass (in g) of the excess reactant is leftover?

3.13 grams is the mass of the excess reactant, which is leftover

Explanation:

This is the reaction.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Let's convert the mass in moles (mass / molar mass)

22 g / 159.7 g / m = 0.137 moles of Fe₂O₃

14.66 g / 28 g/m = 0.523 moles of CO

Ratio is 1:3, 1 mol of oxide react with 3 moles of CO

Then, 0.137 moles will react with (0.137.3) = 0.411 moles of CO

3 mol of CO react with 1 mol of oxide

0.523 moles of CO, would react with (0.523 / 3) = 0.174 moles of Fe₂O₃

So my reactant in excess is the CO, I need 0.411 moles for the reaction and I have 0.523 moles. The limiting is the Fe₂O₃.

Mass of the excess reactant leftover = 0.523 - 0.411 = 0.112 moles

Let's convert the moles in mass (mol. molar mass)

0.112 m. 28 g / m = 3.13 grams