Iron can be obtained by reducing iron (3) oxide with aluminum. Calculate the mass of aluminum and iron oxide (3) required to obtain iron weighing 140 g
m (Fe) = 140 g
m (Al) -?
m (Fe2O3) -?
1) Fe2O3 + 2Al => 2Fe + Al2O3;
2) n (Fe) = m / M = 140/56 = 2.5 mol;
3) n (Al) = n (Fe) = 2.5 mol;
4) m (Al) = n * M = 2.5 * 27 = 67.5 g;
5) n (Fe2O3) = n (Fe) / 2 = 1.25 mol;
6) m (Fe2O3) = n * M = 1.25 * 160 = 200 g.
Answer: The mass of Al is 67.5 g; Fe2O3 – 200 g.
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