We mixed hydrogen with a volume of 8 L and chlorine with a volume of 10 L. The mixture was exploded.

We mixed hydrogen with a volume of 8 L and chlorine with a volume of 10 L. The mixture was exploded. The obtained substances were dissolved in water. Determine the masses of the acids obtained.

Given:
V (H2) = 8 l
V (Cl2) = 10 l

Find:
m (acids) -?

Solution:
1) H2 + Cl2 => 2HCl;
2) n (H2) = V (H2) / Vm = 8 / 22.4 = 0.36 mol;
3) n (Cl2) = V (Cl2) / Vm = 10 / 22.4 = 0.45 mol;
4) n1 (HCl) = n (H2) * 2 = 0.36 * 2 = 0.72 mol;
5) n react. (Cl2) = n (H2) = 0.36 mol;
6) n rest. (Cl2) = n (Cl2) – n re. (Cl2) = 0.45 – 0.36 = 0.09 mol;
7) Cl2 + H2O => HCl + HClO;
8) n2 (HCl) = n rest. (Cl2) = 0.09 mol;
9) n total (HCl) = n1 (HCl) + n2 (HCl) = 0.72 + 0.09 = 0.81 mol;
10) m total. (HCl) = n total (HCl) * M (HCl) = 0.81 * 36.5 = 29.6 g;
11) n (HClO) = n rest. (Cl2) = 0.09 mol;
12) m (HClO) = n (HClO) * M (HClO) = 0.09 * 52.5 = 4.7 g.

Answer: The mass of HCl is 29.6 g; HClO – 4.7 g.