How much chlorine is required to react with 3 moles of aluminum?

1. Let’s compose the equation of the reaction between aluminum and chlorine.

2Al + 3Cl2 = 2AlCl3.

For 2 moles of aluminum, there are 3 moles of chlorine. The substances are in quantitative ratios of 2: 3, that is, the amount of chlorine will be 1.5 times greater than the amount of aluminum.

n mol Cl2 = 0.3 mol × 1.5 = 0.45 mol.

Or:

2 mol Al – 3 mol Cl2,

0.3 mol Al – n mol Cl2.

n mol Cl2 = (0.3 mol × 3 mol): 2 mol = 0.45 mol.

2.Let’s find the volume of chlorine.

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 × 0.45 mol = 10.08 L.

Answer: 10.08 liters.