The volume of carbon monoxide (II) required to reduce iron from 40 g of iron oxide (III) is equal.

| February 15, 2021 | education

1.Calculate the amount of iron oxide (III) substance according to the formula:

n = m: M, where M is molar mass.

M (Fe2O3) = 56 × 2 + 16 × 3 = 160 g / mol.

n = 40: 160 g / mol = 0.25 mol.

2. Let’s make the equation of the reaction of iron reduction by carbon monoxide (II).

Fe2O3 + 3СО = 2Fe + 3СО2 ↑.

According to the reaction equation, there are 3 mol of carbon monoxide (III) per 1 mole of iron oxide (III), that is, the substances are in quantitative ratios of 1: 3, the amount of carbon monoxide (III) substance will be 3 times more than the amount of Fe2O3.

n (Fe2O3) = 3n (CO2) = 0.25 × 3 = 0.75 mol.

3.Find the volume of oxygen

V = Vn n, where Vn is the molar volume of gas equal to 22.4 l / mol, and n is the amount of substance.

V = 22.4 L / mol × 0.75 mol = 16.8 L.

Answer: V = 16.8 liters.



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