When 1.68 g of iron was steamed in air, 2.32 g of iron scale Fe3O4 was formed.

When 1.68 g of iron was steamed in air, 2.32 g of iron scale Fe3O4 was formed. How much oxygen is needed for this reaction?

Let’s implement the solution:

In accordance with the condition of the problem, we will write the data:
m = 1.68 g X g -? m = 2.32 g;

3Fe + 2O2 = Fe3O4 – compounds, iron scale was formed.

We make calculations:
M (Fe) = 55.8 g / mol;

M (O2) = 32 g / mol;

M (Fe3O4) = 167.4 g / mol;

Y (Fe) = m / M = 1.68 / 55.8 = 0.03 mol.

Proportion:
0.03 mol (Fe) – X mol (O2);

-3 mol -2 mol from here, X mol (O2) = 0.03 * 2/3 = 0.02 mol.

We find the mass of the original substance:
m (O2) = Y * M = 0.02 * 32 = 0.64 g

5.The sum of the masses of the starting materials:

m (Fe) + m (O2) = m (Fe2O3);

1.68 + 0.64 = 2.32 g

Answer: you need oxygen weighing 0.64 g

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