How much aluminum is required to interact with 6.72 liters of chlorine?

1. We make the equation:
X g -? V = 6.72 liters.
2Al + 3Cl2 = 2AlCl3 – compounds, aluminum chloride was formed;
2. Calculations:
M (Al) = 26.9 g / mol.
M (Cl2) = 71 g / mol.
3. Proportions:
1 mol of gas at normal level – 22.4 liters.
X mol (Cl2) – 6.72 liters. hence, X mol (Cl2) = 1 * 6.72 / 22.4 = 0.3 mol.
X mol (Al) – 0.3 mol (Cl2);
-2 mol – 3 mol from here, X mol (Al) = 2 * 0.3 / 3 = 0.2 mol.
4. Find the mass of the original substance:
m (Al) = Y * M = 0.2 * 26.9 = 5.38 g.
Answer: it took aluminum with a mass of 5.38 g.