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Дано

w(HCl) = 25% = 0.25

n(Zn) = 0.4 моль

n(Mg) = 0.2 моль

Решение

Zn + 2HCl = ZnCl2 + H2

Mg + 2HCl = MgCl2 + H2

n(HCl) = 2n(Zn) + 2n(Mg) = 2 * 0.4 + 2 * 0.2 = 0.8 + 0.4 = 1.2 моль

m(HCl) = n(HCl)M(HCl) = 1.2 * 36.5 = 43.8 г

m(р-ра) = m(HCl) / w(HCl) = 43.8 / 0.25 = 175.2 г

Ответ:  175.2 г