a)
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{11,648}}{{22,4}} = 0,52{\text{ mol = }}{{\text{n}}_C}\)
\({n_{{H_2}O}} = \frac{{14,04}}{{18}} = 0,78{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 1,56{\text{ mol}}\)
\({n_{{O_2}}} = \frac{{14,56}}{{22,4}} = 0,65{\text{ mol}}\)
Bảo toàn \(O\)
\({n_{O{\text{ trong A}}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,52{\text{ mol}}\)
\( \to n_C:n_H:n_O=0,52:1,56:0,52=1:3:1\)
Vậy \(A\) có dạng \((CH_3O)_n\)
\( \to {M_A} = (12 + 3 + 16)n = 31n = 31{M_{{H_2}}} = 31.2 = 62\)
\( \to n=2 \to A: C_2H_6O_2\)
b)
Sơ đồ phản ứng:
\(B + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{5,824}}{{22,4}} = 0,26{\text{ mol = }}{{\text{n}}_C}\)
\({n_{{H_2}O}} = \frac{{7,02}}{{18}} = 0,39{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,78{\text{ mol}}\)
\({n_{{O_2}}} = \frac{{7,28}}{{22,4}} = 0,325{\text{ mol}}\)
Bảo toàn \(O\)
\({n_{O{\text{ trong B}}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,26{\text{ mol}}\)
\( \to n_C:n_H:n_O=0,26:0,78:0,26=1:3:1\)
Vậy \(B\) có dạng \((CH_3O)_n\)
\( \to {M_B} = (12 + 3 + 16)n = 31n = 31{M_{{H_2}}} = 31.2 = 62\)
\( \to n=2 \to B: C_2H_6O_2\)
