How much chlorine is required for the interaction of 5.4 g of aluminum?

To solve the problem, write down the equation:
m = 5.4 g. X l. -?
1. 2Al + 3Cl2 = 2AlCl3 – the reaction of the compound, aluminum chloride was obtained;
2. Let’s make the calculations:
M (Al) = 26.9 g / mol;
Y (Al) = m / M = 5.4 / 26.9 = 0.2 mol.
3. Proportion:
0.2 mol (Al) – X mol (Cl2);
-2 mol – 3 mol from here, X mol (Cl2) = 0.2 * 3/2 = 0.3 mol.
4. Find the volume of the original substance:
V (Cl2) = 0.3 * 22.4 = 6.72 liters.
Answer: for the reaction, chlorine with a volume of 7.72 liters is required.