Предмет: Химия, автор: artem162721q

К 300г раствора содержащего 0.1 массовой доли гидроксида кальция прилили азотную кислоту, сколько граммов соли образовалось?

Ответы

Автор ответа: BeHappian

Ответ:

примерно 90г

Объяснение:

Решение:

Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O

m(Ca(OH)2) = m(р-ра) * w(Ca(OH)2) = 300 * 0.1 = 30г

n(Ca(OH)2) = m(Ca(OH)2)/MCa(OH)2 = 30/74 = 0.4 моль

0.4/1 = x/1, где х = n(Ca(NO3)2)

n(Ca(NO3)2) = 0.4 моль

m(Ca(NO3)2) = n(Ca(NO3)2) * M(Ca(NO3)2) = 0.4 * 226 = 90.4 г

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