Question

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What masses of iron(III) oxide and aluminum must be used to produce \(15.0 \mathrm{~g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

Short answer

To produce 15.0 g of iron, 13.6 g of iron(III) oxide and 16.20 g of aluminum must be used. The maximum mass of aluminum oxide that could be produced is 13.7 g.

Step-by-step solution

Convert mass of iron to moles

Since the desired amount of iron is 15.0 g, we first need to convert that mass to moles using the molar mass of iron, which is approximately 55.85 g/mol. Moles of iron = \(\frac{15.0 \ \text{g}}{55.85 \ \text{g/mol}}\)

Determine moles of reactants and products using stoichiometry

From the balanced chemical equation, we can see the stoichiometry of the reaction: \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2\mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) For every 2 moles of iron produced, 1 mole of iron(III) oxide and 2 moles of aluminum are required (also 1 mole of aluminum oxide is formed). We can now calculate the moles of iron(III) oxide, aluminum, and aluminum oxide using the stoichiometry and the moles of iron obtained in step 1. Moles of iron(III) oxide = \(\frac{1}{2}\) moles of iron Moles of aluminum = moles of iron Moles of aluminum oxide = \(\frac{1}{2}\) moles of iron

Convert moles of reactants and products to mass

Next, we need to convert the moles of iron(III) oxide, aluminum, and aluminum oxide to mass using their respective molar masses. The molar masses are: - Iron(III) oxide (Fe2O3): 159.69 g/mol - Aluminum (Al): 26.98 g/mol - Aluminum oxide (Al2O3): 101.96 g/mol Mass of iron(III) oxide = moles of iron(III) oxide × molar mass of iron(III) oxide Mass of aluminum = moles of aluminum × molar mass of aluminum Mass of aluminum oxide = moles of aluminum oxide × molar mass of aluminum oxide After calculating the masses using the molar masses, we get the required masses of iron(III) oxide and aluminum to produce 15.0 g of iron, and the maximum mass of aluminum oxide that could be produced.