How much chlorine is needed to react with potassium bromide to obtain 16 g of bromine?

Let’s find the amount of bromine substance Br2.

n = m: M.

M (Br2) = 160 g / mol.

n = 16 g: 160 g / mol = 0.1 mol.

Let’s find the quantitative ratios of substances.

2KBr + Cl2 = 2KCl + Br2

For 1 mole of Br2, there is 1 mole of Cl2.

The substances are in quantitative ratios of 1: 1.

The amount of substances will be equal.

n (Cl2) = n (Br2) = 0.1 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 = 0.1 mol × 22.4 L / mol = 2.24 L.

Answer: 2.24 l.