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Đáp án:

 

Giải thích các bước giải:

`n_{H_2}=3,36÷22,4=0,15 mol`

`Ca + 2H_2O \to Ca(OH)_2 + H_2`

0,15                       0,15 <—0,15

`CaO + H_2O \to Ca(OH)_2`

`=> n_{Ca}=0,15 mol`

`=> m_{CaO}=17,2-0,15.40=11,2g`

`%m_{CaO}={11,2}/{17,2}.100=65,1%`

`%m_{Ca}=100%-65,1%=34,9%`

`n_{CaO}=11,2÷56=0,2 mol`

Theo pt:

`n_{Ca(OH)_2}=nCa+ nCaO`

           `=0,15 + 0,2=0,35 mol`

`=> CM Ca(OH)_2={0,35}/{0,1}=3,5M`

`m_{dd.sau.pư}=17,2+100-0,15.2=116,9g`

`=> C%Ca(OH)_2={0,35.74}/{116,9}.100=22,2%`

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:

\({m_{Ca,CaO}} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 17,2 + 100 = {m_{dd}} + 0,15.2 \to {m_{dd}} = 116,9{\text{ gam}}\)

\({m_{Ca{{(OH)}_2}}} = 0,35.74 = 25,9{\text{ gam}}\)

\( \to \% {m_{Ca{{(OH)}_2}}} = \frac{{25,9}}{{116,9}} = 22,16\% \)