Answer:
229 g Al₂O₃; 243 g Fe₂O₃
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.
M_r: 26.98 159.69 101.96
2Al + Fe₂O₃ ⟶ Al₂O₃ + 2Fe
Mass/g: 121 601
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Step 2. Calculate the moles of each reactant
Moles of Al = 121 × 1/26.98
Moles of Al = 4.485 mol Al
Moles of Fe₂O₃ = 601× 1/159.69
Moles of Fe₂O₃ = 3.764 mol Fe₂O₃
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Step 3. Identify the limiting reactant
Calculate the moles of Al₂O₃ we can obtain from each reactant.
From Al
The molar ratio is 1 mol Al₂O₃:2 mol Al
Moles of Al₂O₃ = 4.485 × 1/2
Moles of Al₂O₃ = 2.242 mol Al₂O₃
From Fe₂O₃:
The molar ratio is 1 mol Al₂O₃:1 mol Fe₂O₃
Moles of Al₂O₃ = 3.764 × 1/1
Moles of Al₂O₃ = 3.764 mol Al₂O₃
The limiting reactant is Al because it gives the smaller amount of Al₂O₃.
The excess reactant is Fe₂O₃.
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Step 4. Calculate the mass of Al₂O₃ formed
Mass of Al₂O₃ = 2.242 × 101.96
Mass of Al₂O₃ = 229 g Al₂O₃
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Step 5. Calculate the moles of Fe₂O₃ reacted
The molar ratio is 1 mol Fe₂O₃:2 mol Al:
Moles of Fe₂O₃ = 4.485 × ½
Moles of Fe₂O₃ = 2.242 mol Fe₂O₃
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Step 6. Calculate the moles of Fe₂O₃ remaining
Moles remaining = original moles – moles used
Moles remaining = 3.764 – 2.242
Moles remaining = 1.521 mol Fe₂O₃
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Step 7. Calculate the mass of Fe₂O₃ remaining
Mass of Fe₂O₃ = 1.521 × 159.69/1
Mass of Fe₂O₃ = 243 g Fe₂O₃
