How much by weight hydrochloric acid will be obtained if 11.2 liters of hydrogen with chlorine gas

How much by weight hydrochloric acid will be obtained if 11.2 liters of hydrogen with chlorine gas entered into the reaction (acid yield 95%)

Given:
V (H2) = 11.2 L;
w (yield. HCl) = 95%;
Find: m (practical HCl) -?
Solution:
1) First, let’s compose the reaction equation:
H2 + Cl2 = 2HCl;
The amount of substance according to the equation: v (H2) = 1 mol, v (Cl2) = 1 mol, v (HCl) = 2 mol;
2) Find v (H2), v = V / Vm, v (H2) = 11.2 L / 22.4 L / mol = 0.5 mol.
3) Let’s make the proportion to find v (HCl):
0.5 mol / 1 mol = v (HCl) / 2 mol;
v (HCl) = 1 mol;
4) Find m (theoretical output) = 1 mol * 36.5 g / mol = 36.5 g
5) w = m (n) / m (t) * 100%
m (n) = 36.5 * 0.95 = 34.675 g.