How much nitrogen is required to interact with 5.4 g of aluminum?

To solve, we compose the reaction equation:
N2 + 2Al = 2AlN – the reaction of the compound, aluminum nitride was obtained:
Let’s make calculations according to the formulas of substances:
M (N2) = 28 g / mol;
M (Al) = 26.9 g / mol;
Let us determine the number of moles of aluminum, if the mass is known:
Y (Al) = m / M = 5.4 / 26.9 = 0.2 mol;
Based on the proportion, we determine the amount of mol of nitrogen:
0.2 mol (Al) – X mol (N2);
-2 mol -1 mol hence, X mol (N2) = 0.2 * 1/2 = 0.1 mol;
Let’s calculate the volume of nitrogen:
V (N2) = 0.1 * 22.4 = 2.24 liters.
Answer: to carry out the reaction, nitrogen with a volume of 2.24 liters is required.