12.5 L of hydrogen chloride was dissolved in 500 ml of water. Determine the percentage

12.5 L of hydrogen chloride was dissolved in 500 ml of water. Determine the percentage of the resulting solution. What reaction of the environment will the solution have if 0.45 mol of sodium hydroxide is added to it?

Given:
V (HCl) = 12.5 l
V (H2O) = 500 ml
n (NaOH) = 0.45 mol

To find:
ω (HCl) -?
the reaction of the environment -?

Decision:
1) n (HCl) = V (HCl) / Vm = 12.5 / 22.4 = 0.558 mol;
2) m (HCl) = n (HCl) * M (HCl) = 0.558 * 36.5 = 20.367 g;
3) m (H2O) = ρ (H2O) * V (H2O) = 1 * 500 = 500 g;
4) m solution (HCl) = m (HCl) + m (H2O) = 20.367 + 500 = 520.367 g;
5) ω (HCl) = m (HCl) * 100% / m solution (HCl) = 20.367 * 100% / 520.367 = 3.91%;
6) HCl + NaOH => NaCl + H2O;
7) n react. (HCl) = n (NaOH) = 0.45 mol;
8) n rest. (HCl) = n (HCl) – n re. (HCl) = 0.558 – 0.45 = 0.108 mol;
9) The reaction of the medium is acidic.

Answer: Mass fraction of HCl is 3.91%; the reaction of the medium is acidic.