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

| May 4, 2021 | education

Let’s find the amount of the substance of iron oxide (III) Fe2O3 by the formula:

n = m: M.

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

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

Let’s compose the reaction equation, find the quantitative ratios of substances.

Fe2O3 + 3CO = 2Fe + 3CO2

According to the reaction equation, there is 3 mol of carbon monoxide per 1 mole of iron oxide. Substances are in quantitative ratios of 1: 3.

The amount of carbon monoxide substance will be 3 times more than the amount of iron oxide substance.

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

Find the volume of carbon monoxide.

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 = 0.75 mol × 22.4 L / mol = 16.8 L.

Answer: 16.8 liters.



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