Question

Al + Fe₃O₄ → Fe + Al₂O₃

Determine the masses of all the substances present after the reaction if  187 g of Al and  791 g of Fe₃O₄ react to completion. Some amounts may be zero (0).

Answer 1

8Al + 3Fe₃O₄ → 9Fe + 4Al₂O₃

M(Al) = 26.98 g/mol

$n(Al) = \frac{187}{26.98} = 6.931 \;mol$

M(Fe₃O₄) = 231.53 g/mol

n(Fe₃O₄) $= \frac{791}{231.53} = 3.416 \;mol$

According to the reaction equation for each 3 mol of Fe₃O₄ we need 8 mol of Al, but we have only 3.416 mol of Fe₃O₄ for 3.416 mol of Al. So, Fe₃O₄ is a limiting reactant.

According to reaction:

From one mole of Fe₃O₄ we can obtain 3 mol of Fe, so

n(Fe) =3n(Fe₃O₄) $= 3 \times 3.416 = 10.249 \;mol$

M(Fe) = 55.84 g/mol

m(Fe) $= 55.84 \times 10.249 = 572.31 \;g$

From 3 mol of Fe₃O₄ we can obtain 4 mol of Al₂O₃, so

n(Al₂O₃) $= \frac{4}{3}n(Fe_3O_4) = \frac{4}{3} \times 3.416 = 4.554 \;mol$

M(Al₂O₃) = 101.96 g/mol

m(Al₂O₃) $= 101.96 \times 4.554 = 464.39 \;g$