Question

ычислите массовую долю (%) продукта, полученного растворением 0,84 г оксида кальция в 200 мл воды.
ответ 0,55

Answer 1

Дано:

$m(H_{2}O)=200gram\\m(CaO)=0.84gram\\\\\omega(Ca(OH)_{2})-?$

Решение:

$CaO+H_{2}O\rightarrow Ca(OH)_{2}\\\\\\\n(CaO)=\frac{m(CaO)}{M(CaO)}=\frac{0.84gram}{56\frac{gram}{mol}}=0.015mol\\n(Ca(OH)_{2})=n(CaO)\\m(Ca(OH)_{2})=n(Ca(OH)_{2})\times M(Ca(OH)_{2})=0.015mol\times 74\frac{gram}{mol}=1.11gram \\\\m'_{p-pa}=m(H_{2}O)+m(CaO)=200gram+0.84gram=200.84gram\\\\\omega(Ca(OH_{2}))=\frac{m(Ca(OH)_{2})}{m'_{p-pa}}=\frac{1.11gram}{200.84gram}=55\times 10^{-3}$

Ответ:

$\omega(Ca(OH)_{2})=0.55\%$.