There are 4.8 moles of Al2O3 produced when 2.4 mol of Fe2O3 reacts with iron.
a)The equation of the reaction is; Fe2O3 (s) + 2Al (s) → 2Fe (l) + Al2O3 (s)
From the reaction equation;
2 moles of Al produces 1 moles of Al2O3
x moles of Al produces 2.4 moles of Al2O3
x = 2 moles × 2.4 moles/1 moles
x = 4.8 moles
b) 1 mole of Fe2O3 produces 1 moles of Al2O3
4.5 moles of Fe2O3 produces 4.5 moles of Al2O3
c) Number of moles in 3.6 g of Al = 3.6 g/27 g/mol = 0.133 moles
If 2 moles of Al yields 1 mole of Al2O3
0.133 moles Al yields 0.133 moles × 1 mole/ 2 moles
= 0.0665 moles
Mass of Al2O3 = 0.0665 moles × 102g/mol
= 6.8 g
d) Number of moles of iron = 3.6 g/56 g/mol = 0.064 moles
From the reaction equation;
1 mole of Al2O3 is produced when 2 moles of iron is produced
x moles of Al2O3 is produced when 0.064 moles is produced
x = 1 mole × 0.064 moles/ 2 moles
x = 0.032 moles
Mass of Al2O3 produced = 0.032 moles × 102g/mol
= 3.3 g
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