Step 1
The reaction must be completed and balanced first.
4 Al + 3 O2 => 2 Al2O3
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Step 2
Information provided:
25.0 g of Al
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Information needed:
The molar mass of Al = 27.0 g/mol (use your periodic table)
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Step 3
By stoichiometry:
Procedure,
4 Al + 3 O2 => 2 Al2O3
4 x 27.0 g Al ------------ 2 x 1 mole Al2O3
25.0 g Al ------------- X
X = 25.0 g Al x 2 x 1 mole Al2O3/4 x 27.0 g Al
X = 1.85 moles
Answer: 1.85 moles of Al2O3