Дано
m = 31.2 г
w(BaCl2) = 20% = 0.2
Решение
m(BaCl2) = mw(BaCl2) = 31.2 * 0.2 = 6.24 г
n(BaCl2) = m(BaCl2) / M(BaCl2) = 6.24 / (137 + 2 * 35.5) = 6.24 / 208 = 0.03 моль
CuSO4 + BaCl2 = BaSO4 + CuCl2
n(BaSO4) = n(BaCl2) = 0.03 моль
m(BaSO4) = n(BaSO4)M(BaSO4) = 0.03 * (137 + 32 + 4 * 16) = 0.03 * 233 = 6.99 г
Ответ: 6.99 г