What mass of salt is formed by the interaction of 50 g of 20% nastrium hydroxide (NaOH) solution with hydrochloric acid?

univerkov | December 3, 2020 | education

NaOH + HCl = NaCl + H2O
m (NaOH) = w * m (solution) = 50 * 0.2 = 10 g
n (NaOH) = m \ M = 10 \ 40 = 0.25 mol
n (NaCl) = n (NaOH) at ur-th p-i = 0.25 mol
m (NaCl) = nM = 0.25 * 58.5 = 14.625 g
Answer: 14.625 g

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