How many liters of hydrogen will be obtained by the interaction of 4 mol of hydrochloric acid with zinc?

One of the ways to solve the problem:
Given:
n (HCL) = 4 mol
To find:
V (H2) =?
1. First you need to compose the reaction equation and equalize it:
2HCl + Zn = ZnCl2 + H2 ↑
2. Using the equation, calculate n hydrogen by the formula. It is known from the condition: n (HCl) = 4 mol, but before hydrochloric acid there is a coefficient equal to two, i.e. HCl is twice the amount of hydrogen. Therefore, n (H2) = 2 mol.
3. Find the volume through the volume under normal conditions:
V (H2) = 22.4 l / mol * 2 mol = 44.8 liters