Answer:
229 g Al₂O₃; 243 g Fe₂O₃
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem. Â
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved. Â
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them. Â
M_r: Â Â Â 26.98 Â Â 159.69 Â Â 101.96
       2Al  +  Fe₂O₃ ⟶ Al₂O₃ + 2Fe
Mass/g: Â 121 Â Â Â Â Â 601
===============
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
===============
Step 3. Identify the limiting reactant Â
Calculate the moles of Alâ‚‚O₃ we can obtain from each reactant. Â
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.
===============
Step 4. Calculate the mass of Al₂O₃ formed
Mass of Al₂O₃ = 2.242 × 101.96
Mass of Al₂O₃ = 229 g Al₂O₃
===============
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
===============
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
==============
Step 7. Calculate the mass of Fe₂O₃ remaining
Mass of Fe₂O₃ = 1.521 × 159.69/1
Mass of Fe₂O₃ = 243 g Fe₂O₃
