1) Balance the chemical reaction.
[tex]Fe_2O_{3(s)}+2Al_{(s)}\rightarrow2Fe_{(l)}+Al_2O_{3(s)}[/tex]
2) List the known and unknown quantities.
Product: Iron (Fe).
Mass: 64.37 g.
Molecular mass: 55.845 g/mol.
Reactant: Aluminum (Al).
Mass: unknown.
Molecular mass: 26.982 g/mol.
3) Convert the mass of Fe to moles of Fe.
[tex]mol\text{ }Fe=64.37\text{ }g\text{ }Fe*\frac{1\text{ }mol\text{ }Fe}{55.845\text{ }g\text{ }Fe}=1.15265\text{ }mol\text{ }Fe[/tex]
4) Convert moles of Fe to moles of Al
The molar ratio between Fe and Al is 2 mol Fe: 2 mol Al.
[tex]mol\text{ }Al=1.15265\text{ }mol\text{ }Fe*\frac{2\text{ }mol\text{ }Al}{2\text{ }mol\text{ }Fe}=1.15265\text{ }mol\text{ }Al[/tex]
5) Convert moles of Al to mass of Al.
[tex]g\text{ }Al=1.15265\text{ }mol\text{ }Al*\frac{26.982\text{ }g\text{ }Al}{1\text{ }mol\text{ }Al}=31.10080\text{ }g\text{ }Al[/tex]
The mass of Aluminum that must be used is 31.10 g Al.
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