<h3>
Answer:</h3>
1.452 g
<h3>
Explanation:</h3>
We are given;
- Mass of Fe₂O₃ as 2.112 g
- Mass of Aluminium as 0.687 g
We are required to calculate the mass of Fe produced;
<h3>Step 1: Determine the number of moles of each reactant</h3>
Moles = Mass ÷ Molar mass
Molar mass of Fe₂O₃ is 159.69 g/mol
Moles of Fe₂O₃ = 2.112 g ÷ 159.69 g/mol
= 0.0132 moles
= 0.013 moles
Molar mass of Aluminium is 26.98 g/mol
Moles of Al = 0.687 g ÷ 26.98 g/mol
= 0.0255 moles
= 0.026 moles
<h3>Step 2: Determine the rate limiting reactant.</h3>
The equation for the reaction is given by;
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
From the reaction, 0.0255 moles of Al would require 0.026 moles × 2 (0.052 moles) of Fe₂O₃.
On the other hand, 0.013 moles of Fe₂O₃ would require 0.026 moles of Aluminium
Therefore, Fe₂O₃ is the rate limiting reactant for the reaction.
<h3>
Step 3: Moles of Fe produced </h3>
1 mole of Fe₂O₃ reacts to produce 2 moles of Fe
Therefore; Moles of Fe = Moles of Fe₂O₃ × 2
= 0.013 moles × 2
= 0.026 Moles
<h3>Step 4: Mass of Fe produced </h3>
Mass = Number of moles × Molar mass
Molar mass of Fe is 55.845 g/mol
Therefore;
Mass of Fe = 0.026 moles × 55.845 g/mol
= 1.452 g
Therefore, 1.452 g of Fe were produced.
