Calculate the volume of carbon monoxide 2 required to completely reduce 10 g of iron oxide 3.

univerkov | April 7, 2021 | education

Data: mFe – mass of iron oxide (mFe = 10 g).

Const: Vm – molar volume (for standard service Vm = 22.4 l / mol); МFe2O3 – molar mass of iron oxide (МFe2O3 = 159.69 g / mol ≈ 160 g / mol).

1) Equation: 2Fe (iron) + 3СО2 (carbon dioxide) = Fe2O3 (iron oxide) + 3СО (carbon monoxide).

2) Amount of substance:

Iron oxide: νFe2O3 = mFe2O3 / МFe2O3 = 10/160 ≈ 0.0625 mol.

Carbon monoxide: νСО / νFe2O3 = 3/1 and νСО = 3 * νFe2O3 = 3 * 0.0625 = 0.1875 mol.

3) The required volume of carbon monoxide: VСО = νСО * Vm = 0.1875 * 22.4 = 4.2 liters.

Answer: You need to use 4.2 liters of carbon monoxide.

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