Step 1
The reaction involved here:
3 FeO + 2 Al => 3 Fe + AI2O3 (Don't forget to completed it and balance it)
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Step 2
Data provided:
125 grams of FeO react with 25.0 grams of AI
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Data needed:
The molar masses of:
FeO) 71.8 g/mol
Al) 27.0 g/mol
Fe) 55.8 g/mol
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Step 3
The limiting reactant:
By stoichiometry,
3 FeO + 2 Al => 3 Fe + AI2O3
3 x 71.8 g FeO ----- 2 x 27.0 g Al
125 g FeO ----- X
X = 125 g FeO x 2 x 27.0 g Al/3 x 71.8 g FeO
X = 31.3 g Al
For 125 g of FeO, 31.3 g of Al is needed, but there is 25.0 g of Al, so the limiting reactant is Al.
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Step 4
The amount of Fe produced:
By stoichiometry,
3 FeO + 2 Al => 3 Fe + AI2O3
2 x 27.0 g Al ---- 3 x 55.8 g Fe
25.0 g Al --- X = 77.5 g Fe
Answer: 77.6 g Fe (the closest value)