Calculate the mass of iron that is formed by the interaction of 20 g of iron (III) oxide with 50 g of aluminum.


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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