Calculate the volume of hydrogen released by the interaction of 40.5 g of aluminum with a sulfuric acid solution.

Let’s write the reaction equation:

2Al + 3H2SO4 = Al2 (SO) 4 + 3H2

Let’s find the amount of aluminum:

n (Al) = m (Al) / M (Al) = 40.5 g / 27 g / mol = 1.5 mol

Let’s find the amount of released hydrogen:

n (H2) = 1.5 * n (Al) = 1.5 * 1.5 mol = 2.25 mol

Let’s calculate the volume of hydrogen:

V (H2) = n (H2) * Vm = 2.25 mol * 22.4 L / mol = 50.4 L

Answer: the volume of hydrogen is 50.4 liters.