How much chlorine in liters is required for the reaction with 9.3 grams of iron.

Let us find the amount of Fe by the formula:

n = m: M.

M (Fe) = 56g / mol.

n = 9.3 g: 56 g / mol = 0.166 mol.

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

2Fe + 3Cl2 = 2FeCl3.

According to the reaction equation, there is 3 mol of Cl2 for 2 mol of Fe. Substances are found in quantitative ratios 2: 3 = 1: 1.5. The amount of Cl2 is 1.5 times greater than the amount of Fe.

n (Cl2) = 1.5n (Fe) = 0.166 × 1.5 mol = 0.249 mol.

Find the volume of Cl2.

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 = 0.249 mol × 22.4 L / mol = 5.58 L.

Answer: 5.58 liters.