Đáp án:
\(CH_3CH_2CH_2CH_3\);
\(CH_3CH(CH_3)CH_3\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(X + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \dfrac{{10,752}}{{22,4}} = 0,48{\text{ mol;}}\\{{\text{n}}_{{H_2}O}} = \dfrac{{10,8}}{{18}} = 0,6{\text{ mol}}\)
Bảo toàn khối lượng:
\({m_X} + {m_{{O_2}}} = {m_{C{O_2}}} + {m_{{H_2}O}}\)
\( \to 6,96 + {m_{{O_2}}} = 0,48.44 + 10,8\)
\( \to {m_{{O_2}}} = 24,96{\text{ gam}}\)
\( \to {n_{{O_2}}} = \dfrac{{24,96}}{{32}} = 0,78{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,78.22,4 = 17,472{\text{ lít}}\)
Ta có:
\({n_C} = {n_{C{O_2}}} = 0,48{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 1,2{\text{ mol}}\)
\( \to {n_O} = \frac{{6,96 - 0,48.12 - 1,2.1}}{{16}} = 0\)
Vậy \(X\) chỉ chứa \(C;H\)
\({n_{{H_2}O}} > {n_{C{O_2}}} \to X:ankan \\\to {n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,12{\text{ mol}}\)
\( \to {C_X} = \dfrac{{0,48}}{{0,12}} = 4\)
Suy ra \(X\) là \(C_4H_{10}\)
CTCT:
\(CH_3CH_2CH_2CH_3\); \(CH_3CH(CH_3)CH_3\)
