The volume of hydrogen formed by the interaction of 4.8 magnesium with an excess of hydrochloric acid is equal to what?

Let’s implement the solution:

According to the condition of the problem, we compose the equation of the process:
m = 4.8 g Chl -?

Mg + 2HCl = MgCl2 + H2 – ОВР, hydrogen was obtained;

We make calculations:
M (Mg) = 24.3 g / mol;

M (H2) = 2 g / mol.

Determine the amount of the original substance:
Y (Mg) = m / M = 4.8 / 24.3 = 0.2 mol;

Y (H2) = 0.2 mol since the amount of substances is 1 mol.

Find the volume of the product:
V (H2) = 0.2 * 22.4 = 4.48 L

Answer: the volume of hydrogen is 4.48 liters.