Chemistry - limiting reagent

Question

Given the following equation:

Al2O3 + Fe → Fe3O4 + Al

1. Balance the equation.
2. If 25.4 g of Al2O3 is reacted with 10.2 g of Fe, determine the limiting reagent.
3. Determine the number of moles of Al produced.
4. Determine the number of grams of Fe3O4 produced.
5. Determine the number of grams of excess reagent left over in the reaction.

Answer 1

1). 4Al₂O₃ + 9Fe → 3Fe₃O₄ + 8Al ... balanced equation

2). Limiting reactant can be found by dividing moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For Al₂O₃ we have 25.4 g x 1 mol / 102 g = 0.249 mols (÷4->0.06)

For Fe we have 10.2 g x 1 mol / 55.9 g = 0.182 mols (÷9->0.02)

Since 0.02 is less than 0.06, Fe is the limiting reactant

3). mols Al produced. This is determined by the mols of the limiting reactant.

0.182 mols Fe x 8 mols Al / 9 mols Fe = 0.162 mols Al

4). grams of Fe₃O₄ produced is also determined by the mols of limiting reactant.

0.182 mols Fe x 3 mols Fe₃O₄ / 9 mols Fe x 232 g Fe₃O₄/mol = 14.1 g Fe₃O₄

5). grams of excess reagent left over. Excess reagent is Al₂O₃.

First, find moles Al₂O₃ used up: 0.182 mol Fe x 4 mol Al₂O₃ / 9 mol Fe = 0.081 mols Al₂O₃ used

Next, subtract this amount from the original moles at the start of the reaction:

mols left over = 0.249 mols - 0.081 mols = 0.168 mols Al₂O₃ left over

converting to grams, we have 0.168 mols x 102 g/mol = 17.1 g Al₂O₃ left over