Calculate the mass fraction of salt in the solution obtained by reacting 200 g of a potassium hydroxide

Calculate the mass fraction of salt in the solution obtained by reacting 200 g of a potassium hydroxide solution with a mass fraction of a solute of 0.056 and 100 g of a solution containing the required amount of hydrogen chloride.

Given:
m solution (KOH) = 200 g
ω (KOH) = 0.056
m solution (HCl) = 100 g

To find:
ω (salt) -?

1) KOH + HCl => KCl + H2O;
2) m (KOH) = ω (KOH) * m solution (KOH) = 200 * 0.056 = 11.2 g;
3) M (KOH) = Mr (KOH) = Ar (K) * N (K) + Ar (O) * N (O) + Ar (H) * N (H) = 39 * 1 + 16 * 1 + 1 * 1 = 56 g / mol;
4) n (KOH) = m (KOH) / M (KOH) = 11.2 / 56 = 0.2 mol;
5) n (KCl) = n (KOH) = 0.2 mol;
6) M (KCl) = Mr (KCl) = Ar (K) * N (K) + Ar (Cl) * N (Cl) = 39 * 1 + 35.5 * 1 = 74.5 g / mol;
7) m (KCl) = n (KCl) * M (KCl) = 0.2 * 74.5 = 14.9 g;
8) m solution (KCl) = m solution (KOH) + m solution (HCl) = 200 + 100 = 300 g;
9) ω (KCl) = m (KCl) / m solution (KCl) = 14.9 / 300 = 0.05.

Answer: Mass fraction of KCl is 0.05.