Đáp án:
\(\% {m_{Ca{{(OH)}_2}}} = 36,156\% ; \% {m_{CaO}} = 63,844\% \)
\( C{\% _{CaC{l_2}}} = 9,56\% \)
Giải thích các bước giải:
Giả sử ban đầu có 1 mol \(CaO\)
Phản ứng xảy ra:
\(CaO + {H_2}O\xrightarrow{{}}Ca{(OH)_2}\)
Ta có:
\({m_{CaO}} = 1.(40 + 16) = 56{\text{ gam}}\)
\( \to {m_E} = 56.109,65\% = 61,4{\text{ gam}}\)
\( \to {m_{{H_2}O}} = {m_E} - {m_{CaO}} = 5,4{\text{ gam}}\)
\( \to {n_{{H_2}O}} = \frac{{5,4}}{{18}} = 0,3{\text{ mol = }}{{\text{n}}_{Ca{{(OH)}_2}}}\)
\( \to {n_{CaO{\text{ dư}}}} = 1 - 0,3 = 0,7{\text{ mol}}\)
\({m_{Ca{{(OH)}_2}}} = 0,3.(40 + 17.2) = 22,2{\text{ gam}}\)
\( \to \% {m_{Ca{{(OH)}_2}}} = 36,156\% ; \% {m_{CaO}} = 63,844\% \)
61,4 gam \(E\) chứa 0,3 mol \(Ca(OH)_2\) và 0,7 mol \(CaO\)
Vậy 12,28 gam \(E\) chứa
\({n_{Ca{{(OH)}_2}}} = 0,3.\frac{{12,28}}{{61,4}} = 0,06{\text{ mol}}\)
\({n_{CaO}} = 0,7.\frac{{12,28}}{{61,4}} = 0,14{\text{ mol}}\)
\(Ca{(OH)_2} + 2HCl\xrightarrow{{}}CaC{l_2} + 2{H_2}O\)
\(CaO + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O\)
\({m_{dd\;{\text{HCl}}}} = 200.1,1 = 220{\text{ gam}}\)
\( \to {m_{HCl}} = 220.20\% = 44{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{44}}{{36,5}} > 2{n_{CaO}} + 2{n_{Ca{{(OH)}_2}}}\)
BTKL:
\({m_{dd}} = {m_E} + {m_{dd{\text{ HCl}}}} = 12,28 + 220 = 232,28{\text{ gam}}\)
\({n_{CaC{l_2}}} = {n_{CaO}} + {n_{Ca{{(OH)}_2}}} = 0,06 + 0,14 = 0,2{\text{ mol}}\)
\( \to {m_{CaC{l_2}}} = 0,2.111 = 22,2{\text{ gam}}\)
\( \to C{\% _{CaC{l_2}}} = \frac{{22,2}}{{232,28}} = 9,56\% \)
