Đáp án:

\({C_2}{H_5}OH\)

Giải thích các bước giải:

 Đốt cháy hỗn hợp A:

\({C_6}{H_6} + 7,5{O_2}\xrightarrow{{{t^o}}}6C{O_2} + 3{H_2}O\)

\({C_n}{H_{2n + 1}}OH + 1,5n{O_2}\xrightarrow{{{t^o}}}nC{O_2} + (n + 1){H_2}O\)

Ta có: 

\({n_{{H_2}O}} = \frac{{3,24}}{{18}} = 0,18{\text{ mol}}\)

\({n_{C{O_2}}} = {n_Y} - {n_Z} = 65,744 - 62,16 = 3,584{\text{ lít}} \to {{\text{n}}_{C{O_2}}} = \frac{{3,584}}{{22,4}} = 0,16{\text{ mol}}\)

\({n_C} = {n_{C{O_2}}} = 0,16{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,36{\text{ mol}} \to {{\text{n}}_O} = \frac{{3,08 - 0,16.12 - 0,36.1}}{{16}} = 0,05{\text{ mol = }}{{\text{n}}_X}\)

\({n_{{H_2}O}} - {n_{C{O_2}}} = 0,18 - 0,16 = 0,02{\text{ mol = }}{{\text{n}}_X} - 3{n_{{C_6}{H_6}}} \to {n_{{C_6}{H_6}}} = 0,01{\text{ mol}}\)

\( \to {n_C} = 6.{n_{{C_6}{H_6}}} + n.{n_X} = 0,16 \to n = \frac{{0,16 - 0,01.6}}{{0,05}} = 2\)

Vậy X là \({C_2}{H_5}OH\)

Ta có: 

\({m_{{C_6}{H_6}}} = 0,01.78 = 0,78{\text{ gam}} \to {\text{\% }}{{\text{m}}_{{C_6}{H_6}}} = \frac{{0,78}}{{3,08}} = 25,3\%  \to \% {m_{{C_2}{H_5}OH}} = 74,7\% \)

Ta có: 

\({n_{{O_2}{\text{ phản ứng}}}} = 7,5{n_{{C_6}{H_6}}} + 3{n_{{C_2}{H_5}OH}} = 0,01.7,5 + 3.0,05 = 0,225{\text{ mol}} \to {{\text{V}}_{{O_2}{\text{ phản ứng}}}} = 0,225.22,4 = 5,04{\text{ lítt}}\)

\( \to V = {V_Z} + {V_{{O_2}{\text{ phản ứng}}}} = 62,16 + 5,04 = 67,2{\text{ lítt}}\)