Find the volume of gas released during the interaction of 27 g of hydrochloric acid with aluminum?

1. Let’s find the amount of aluminum substance.

n = m: M

M (Al) = 27 g / mol.

n = 27 g: 27 g / mol = 1 mol.

2. Let’s compose the equation of the reaction between aluminum and hydrochloric acid. Let us determine in what quantitative ratios the substances are – aluminum and hydrogen.

2Al + 6HCl = 2AlCl3 + 3H2.

For 2 moles of aluminum, there are 3 moles of hydrogen, that is, the quantitative ratios are 2: 3 or 1: 1.5.

Then the amount of hydrogen substance is 1.5 more than the amount of aluminum substance.

1 mol × 1.5 = 1.5 mol.

3.The volume is determined by the formula:

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 1.5 mol × 22.4 L / mol = 33.6 L.

Answer: 33.6 liters.