Final answer:
The mass of iron produced when 21.4g of aluminum reacts with 91.4g of Fe2O3 is about 63.89 grams, calculated using stoichiometry and the balanced chemical equation.
Step-by-step explanation:
To determine the mass of iron that will be produced when 21.4g of aluminum reacts with 91.4g of Fe2O3, we must first write the balanced chemical equation:
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
We know that 1 mole of Fe2O3 reacts with 2 moles of Al to produce 2 moles of Fe. The molar masses are approximately 159.7 g/mol for Fe2O3 and 55.85 g/mol for Fe. Using stoichiometry, we can calculate the number of moles of Fe2O3 we have:
(91.4g Fe2O3) / (159.7g/mol Fe2O3) = 0.572 moles of Fe2O3
From the balanced equation, we can see that 1 mole of Fe2O3 produces 2 moles of Fe. Therefore, 0.572 moles of Fe2O3 will produce 0.572 * 2 = 1.144 moles of Fe. Finally, we convert moles of Fe to grams:
1.144 moles of Fe * 55.85 g/mol = 63.89g of Fe
Therefore, the mass of iron produced will be about 63.89 grams, assuming complete reaction and that aluminum is the limiting reactant.