Đáp án:
\(X\) là \(C_2H_6O\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(X + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = {n_C} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)
\({n_{{H_2}O}} = \frac{{5,4}}{{18}} = 0,3{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,6{\text{ mol}}\)
\( \to {n_O} = \frac{{4,6 - 0,2.12 - 0,6.1}}{{16}} = 0,1{\text{ mol}}\)
\({M_X} = 23{M_{{H_2}}} = 23.2 = 46 \to {n_X} = \frac{{4,6}}{{46}} = 0,1\)
\( \to {C_X} = \frac{{0,2}}{{0,1}} = 2;{H_X} = \frac{{0,6}}{{0,1}} = 6;{O_X} = \frac{{0,1}}{{0,1}} = 1\)
Vậy \(X\) là \(C_2H_6O\)
Ta có:
\({n_{CaC{O_3}}} = \frac{{10}}{{100}} = 0,1{\text{ mol < }}{{\text{n}}_{C{O_2}}}\)
Vậy phản ứng xảy ra:
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
\(Ca{(OH)_2} + 2C{O_2}\xrightarrow{{}}Ca{(HC{O_3})_2}\)
\({n_{Ca{{(HC{O_3})}_2}}} = \frac{{0,2 - 0,1}}{2} = 0,05{\text{ mol}}\)
\( \to {n_{Ca{{(OH)}_2}}} = {n_{CaC{O_3}}} + {n_{Ca{{(HC{O_3})}_2}}} = 0,1 + 0,05 = 0,15\)
\( \to {C_{M{\text{ Ca(OH}}{{\text{)}}_2}}} = \frac{{0,15}}{{0,1}} = 1,5M\)
