Answer:
229 g Al₂O₃; 243 g Fe₂O₃
Explanation:
We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. <em>Gather all the information</em> 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
===============
Step 2. Calculate the <em>moles of each reactant </em>
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 <em>limiting reactant</em>
Calculate the moles of Al₂O₃ we can obtain from each reactant.
<em>From Al
</em>
The molar ratio is 1 mol Al₂O₃:2 mol Al
Moles of Al₂O₃ = 4.485 × 1/2
Moles of Al₂O₃ = 2.242 mol Al₂O₃
<em>From Fe₂O₃</em>:
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 <em>limiting reactant</em> is Al because it gives the smaller amount of Al₂O₃.
The <em>excess reactant</em> is Fe₂O₃.
===============
Step 4. Calculate the <em>mass of Al₂O₃ formed
</em>
Mass of Al₂O₃ = 2.242 × 101.96
Mass of Al₂O₃ = 229 g Al₂O₃
===============
Step 5. Calculate the <em>moles of Fe₂O₃ reacted
</em>
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 <em>moles of Fe₂O₃ remaining
</em>
Moles remaining = original moles – moles used
Moles remaining = 3.764 – 2.242
Moles remaining = 1.521 mol Fe₂O₃
==============
Step 7. Calculate the <em>mass of Fe₂O₃ remaining
</em>
Mass of Fe₂O₃ = 1.521 × 159.69/1
Mass of Fe₂O₃ = 243 g Fe₂O₃
