m (Fe2O3) = 20 g
m (Al) = 50 g
m (Fe) -?
1) We compose the reaction equation corresponding to the condition of the problem:
Fe2O3 + 2Al = 2Fe + Al2O3;
2) Find the amount of iron oxide and aluminum:
n (Fe2O3) = m: M = 20 g: 160 g / mol = 0.125 mol
n (Al) = m: M = 50 g: 27 g / mol = 1.85 mol
We carry out calculations, substituting a smaller value in order to get more accurate calculations. We work with Fe2O3:
3) We compose a logical expression:
if 1 mol of Fe2O3 gives 2 mol of Fe in the reaction,
then 0.125 mol of Fe2O3 will give a mol of Fe in the reaction x,
then x = 0.25 mol.
4) Find the mass of iron released during the reaction:
m (Fe) = n * M = 0.25 mol * 56 g / mol = 11.2 g;
Answer: m (Fe) = 11.2 grams.
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