When 5.4 g of solid aluminum (molar mass = 27 g/mol) reacts completely with excess iron(III) oxide via the thermite reaction: 2Al + Fe2O3 -> Al2O3 + 2Fe, what mass of molten iron (molar mass = 56 g/mol) is theoretically produced?
Answer and explanation
Correct answer: A. 11.2 g
- Step 1: Calculate moles of aluminum reacting: n(Al) = 5.4 g / 27 g/mol = 0.20 mol.
- Step 2: Determine stoichiometric mole ratio from the balanced equation: 2 mol Al yields 2 mol Fe, so mole ratio is 1:1.
- Step 3: Calculate moles of iron produced: n(Fe) = 0.20 mol.
- Step 4: Compute theoretical mass of iron: mass = 0.20 mol * 56 g/mol = 11.2 g.
Answer
11.2 g.
Why not the others: '5.6 g' mistakenly assumes a 2:1 ratio halving the iron produced; '22.4 g' doubles the stoichiometric yield; '16.8 g' uses an incorrect molar mass ratio.
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