Question

What is the mass of aluminum metal that reacts to give 1.00 g of iron?

1) __feo(l)
2) __al(l)
3) __fe(l)
4) __Al₂O₃(s)

Answer 1

To produce 1.00 g of iron, 0.483 g of aluminum are required based on stoichiometric calculations from the balanced chemical equation.

Step-by-step explanation:

To find the mass of aluminum that reacts to produce 1.00 g of iron, we must use stoichiometry based on the conversion between the moles of iron and aluminum in the balanced equation:

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s).

Firstly, we find the moles of iron that correspond to 1.00 g:

• (1.00 g Fe) / (55.85 g/mol Fe) = 0.0179 mol Fe

Using the stoichiometric ratio between Fe and Al from the balanced equation, we can find the moles of Al needed:

• (0.0179 mol Fe) × (2 mol Al / 2 mol Fe) = 0.0179 mol Al

Finally, we convert the moles of aluminum to grams using its molar mass (26.98 g/mol Al):

• (0.0179 mol Al) × (26.98 g/mol Al) = 0.483 g Al

Therefore, 0.483 g of aluminum are required to react and produce 1.00 g of iron.