What is the volume of hydrogen released when 5.4 g of aluminum reacts with sulfuric acid?

Reaction equation: 2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2
Let us find the amount of aluminum substance n = m / M = 5.4 g / 27 g / mol = 0.2 mol. (M – molar mass, determined by the periodic table). According to the reaction from 2 mol of aluminum, 3 mol of hydrogen is obtained, we will make the proportion 0.2 / 2 = n (H2) / 3. n (H2) = 0.2 * 3/2 = 0.3 mol. Let’s find the volume of hydrogen V = n * Vm = 0.3 mol * 22.4 l / mol = 6.72 l. (Vm is molar volume, constant for all gases under normal conditions).
Answer: the volume of hydrogen is 6.72 liters.