What volume of chlorine is required to interact with iron weighing 84 g?

Let’s find the amount of iron substance by the formula:

n = m: M.

M (Fe) = 56 g / mol.

n = 84 g: 56 g / mol = 1.5 mol.

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

2Fe + 3Cl2 = 2FeCl3.

According to the reaction equation, for 2 mol of iron there is 3 mol of chlorine. Substances are in quantitative ratios of 2: 3.

The amount of chlorine will be 1.5 times more than the amount of iron.

n (Cl2) = 1.5 n (Fe) = 1.5 × 1.5 = 2.25 mol.

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 × 2.25 mol = 50.4 L.

Answer: 50.4 liters.