Đáp án:
\(\% {m_{Ca}} = 34,88\% ;\% {m_{CaO}} = 65,12\% \)
\({C_{M{\text{ Ca(OH}}{{\text{)}}_2}}}= 3,5M\)
\(\% {m_{Ca{{(OH)}_2}}} = 22,16\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Ca + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {H_2}\)
\(CaO + {H_2}O\xrightarrow{{}}Ca{(OH)_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}{{\text{n}}_{Ca}}\)
\( \to {m_{Ca}} = 0,15.40 = 6{\text{ gam}}\)
\( \to {m_{CaO}} = 17,2 - 6 = 11,2{\text{ gam}}\)
\( \to {n_{CaO}} = \frac{{11,2}}{{40 + 16}} = 0,2{\text{ mol}}\)
\(\% {m_{Ca}} = \frac{6}{{17,2}} = 34,88\% \to \% {m_{CaO}} = 65,12\% \)
Ta có:
\({n_{Ca{{(OH)}_2}}} = {n_{Ca}} + {n_{Ca{{(OH)}_2}}} = 0,15 + 0,2 = 0,35{\text{ mol}}\)
\( \to {C_{M{\text{ Ca(OH}}{{\text{)}}_2}}} = \frac{{0,35}}{{0,1}} = 3,5M\)
\({m_{{H_2}O}} = 100.1 = 100{\text{ gam}}\)
BTKL:
\({m_{Ca,CaO}} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 17,2 + 100 = {m_{dd}} + 0,15.2 \to {m_{dd}} = 116,9{\text{ gam}}\)
\({m_{Ca{{(OH)}_2}}} = 0,35.74 = 25,9{\text{ gam}}\)
\( \to \% {m_{Ca{{(OH)}_2}}} = \frac{{25,9}}{{116,9}} = 22,16\% \)
