How much aluminum is required to react with chlorine if you get 53.4 g of aluminum chloride?

The equation for the reaction between aluminum and chlorine is as follows:

2 Al + 3 Cl2 = 2 AlCl3

Molar mass of aluminum = 27 g / mol;

molar mass of aluminum chloride = 132 g / mol

According to the reaction equation, we make up the proportion:

2 * 27 g / mol of aluminum corresponds to 2 * 132 g / mol of aluminum chloride as

X g of aluminum corresponds to 53.4 g of aluminum chloride.

X = (53.4 * 2 * 27) / (2 * 132) = 10.9g

Answer: 10.9 g of aluminum