Answer: 64.20 g
Step-by-step explanation:
The balanced chemical equation for the reaction between aluminum and iron(III) oxide is:
2Al + Fe2O3 → 2Fe + Al2O3
Using the molar masses of aluminum (Al) and iron(III) oxide (Fe2O3), we can calculate the number of moles of each reactant:
moles of Al = mass ÷ molar mass = 21.4 g ÷ 26.98 g/mol = 0.793 mol
moles of Fe2O3 = mass ÷ molar mass = 91.3 g ÷ 159.69 g/mol = 0.572 mol
According to the balanced equation, 2 moles of Al react with 1 mole of Fe2O3. Therefore, the stoichiometric ratio of Al to Fe2O3 is 2:1. However, we have more moles of Al than what is required for the reaction with the available amount of Fe2O3. Hence, Fe2O3 is the limiting reagent and Al is in excess.
To calculate the amount of excess Al, we can use the stoichiometric ratio of Al to Fe2O3 to determine the theoretical amount of Al required to react with all the available Fe2O3:
moles of Al needed = 0.5 × moles of Fe2O3 = 0.5 × 0.572 mol = 0.286 mol
The amount of excess Al is the difference between the actual amount of Al used and the theoretical amount needed:
moles of excess Al = moles of Al used - moles of Al needed
= 0.793 mol - 0.286 mol
= 0.507 mol
The mass of excess Al can be calculated using its molar mass:
mass of excess Al = moles of excess Al × molar mass of Al
= 0.507 mol × 26.98 g/mol
= 13.68 g
Therefore, the mass of excess Al left over is 13.68 g.
To determine the amount of iron produced, we can use the stoichiometric ratio of Fe2O3 to Fe in the balanced equation:
1 mole of Fe2O3 produces 2 moles of Fe
moles of Fe = 2 × moles of Fe2O3
= 2 × 0.572 mol
= 1.144 mol
The mass of iron produced can be calculated using its molar mass:
mass of Fe = moles of Fe × molar mass of Fe
= 1.144 mol × 55.85 g/mol
= 64.20 g
Therefore, the mass of iron produced is 64.20 g.