Calculate the mass of salt that can be obtained from the interaction of 4.32 g of aluminum with chlorine.

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

n = m: M.

M (Al) = 27 g / mol.

n = 4.32 g: 27 g / mol = 0.16 mol.

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

2Al + 3Cl2 = 2AlCl3.

According to the reaction equation, 2 mol of Al accounts for 2 mol of AlCl3.

Substances are in quantitative ratios 1: 1.

n (AlCl3) = n (Al) = 0.16 mol.

Let’s find the mass of AlCl3 by the formula:

m = n × M,

M (AlCl3) = 133.5 g / mol.

m = 0.16 mol × 133.5 g / mol = 21.36 g.

Answer: 21.36 g.