Hỗn hợp X gồm ankan (a mol), anken, ankin (a mol)....

C

- Hướng dẫn giải

\(\begin{gathered} ({k_{ankan}} - 1).{n_{ankan}} + ({k_{anken}} - 1).{n_{anken}} + ({k_{ankin}} - 1).{n_{ankin}} = {n_{C{O_2}}} - {n_{{H_2}O}} \hfill \\ \Rightarrow - {n_{ankan}} + {n_{ankin}} = {n_{C{O_2}}} - {n_{{H_2}O}} \Rightarrow - a + a = {n_{C{O_2}}} - {n_{{H_2}O}} \Rightarrow {n_{C{O_2}}} = {n_{{H_2}O}} \hfill \\ \Rightarrow \frac{{2b + 5,6}}{{44}} = \frac{b}{{18}} \Rightarrow b = 12,6{\text{ gam}} \Rightarrow {n_{C{O_2}}} - {n_{{H_2}O}} = \frac{{12,6}}{{18}} = 0,7{\text{ mol}} \hfill \\ \end{gathered} \)

\(\begin{gathered} \xrightarrow{{BT{\text{ O}}}}2.{n_{{O_2}}} = 2.{n_{C{O_2}}} + {n_{H{}_2O}} \Rightarrow 2.{n_{{O_2}}} = 2.0,7 + 0,7 \Rightarrow {n_{{O_2}}} = 1,05{\text{ mol}} \hfill \\ {\text{V = }}{{\text{V}}_{{O_2}}} = 1,05.22,4 = 23,52{\text{ li t}} \hfill \\ \end{gathered} \)

\(m = {m_{C(X)}} + {m_{H(X)}} = 12.0,7 + 2.0,7 = 9,8{\text{ gam}}{\text{.}}\)