Дано
m = 50 г
w(KOH) = 20% = 0.2
Решение
m(KOH) = mw(KOH) = 50 * 0.2 = 10 г
n(KOH) = m(KOH) / M(KOH) = 10 / (39 + 16 + 1) = 10 / 56 = 0.18 моль
2KOH + CuSO4 = Cu(OH)2 + K2SO4
n(Cu(OH)2) = 0.5n(KOH) = 0.5 * 0.18 = 0.09 моль
m(Cu(OH)2) = n(Cu(OH)2)M(Cu(OH)2) = 0.09 * (64 + 2 * 16 + 2 * 1) = 0.09 * 98 = 8.82 г
Ответ: 8.82 г
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Answers & Comments
Дано
m = 50 г
w(KOH) = 20% = 0.2
Решение
m(KOH) = mw(KOH) = 50 * 0.2 = 10 г
n(KOH) = m(KOH) / M(KOH) = 10 / (39 + 16 + 1) = 10 / 56 = 0.18 моль
2KOH + CuSO4 = Cu(OH)2 + K2SO4
n(Cu(OH)2) = 0.5n(KOH) = 0.5 * 0.18 = 0.09 моль
m(Cu(OH)2) = n(Cu(OH)2)M(Cu(OH)2) = 0.09 * (64 + 2 * 16 + 2 * 1) = 0.09 * 98 = 8.82 г
Ответ: 8.82 г