# Calculate the mass of the salt formed by the interaction of 150 g of 20% hydrochloric

Calculate the mass of the salt formed by the interaction of 150 g of 20% hydrochloric acid solution with a sufficient amount of sodium hydroxide.

Given:
m solution (HCl) = 150 g
ω (HCl) = 20%

Find:
m (salt) -?

1) HCl + NaOH => NaCl + H2O;
2) m (HCl) = ω (HCl) * m solution (HCl) / 100% = 20% * 150/100% = 30 g;
3) M (HCl) = Mr (HCl) = Ar (H) * N (H) + Ar (Cl) * N (Cl) = 1 * 1 + 35.5 * 1 = 36.5 g / mol;
4) n (HCl) = m (HCl) / M (HCl) = 30 / 36.5 = 0.82 mol;
5) n (NaCl) = n (HCl) = 0.82 mol;
6) M (NaCl) = Mr (NaCl) = Ar (Na) * N (Na) + Ar (Cl) * N (Cl) = 23 * 1 + 35.5 * 1 = 58.5 g / mol;
7) m (NaCl) = n (NaCl) * M (NaCl) = 0.82 * 58.5 = 48 g.

Answer: The mass of NaCl is 48 g.

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