Giải thích các bước giải:

1,

\(\begin{array}{l}
{n_{C{O_2}}} = 0,1mol\\
{n_{{H_2}O}} = 0,125mol\\
{n_{{H_2}O}} > {n_{C{O_2}}}
\end{array}\)

Hidrocacbon X là ankan

\(\begin{array}{l}
{n_{C{O_2}}} = 0,1mol\\
{n_{{H_2}O}} = 0,125mol\\
{n_{{H_2}O}} > {n_{C{O_2}}}\\
{C_n}{H_{2n + 2}} + \dfrac{{3n + 1}}{2}{O_2} \to nC{O_2} + (n + 1){H_2}O\\
{n_{ankan}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,025mol\\
C = \dfrac{{{n_{C{O_2}}}}}{{{n_{ankan}}}} = 4\\
{C_4}{H_{10}}
\end{array}\)

\(C{H_3} - {\rm{ }}C{H_2} - {\rm{ }}C{H_2} - {\rm{ }}C{H_3} =  > {\rm{ }}Butan\)

\(C{H_3} - {\rm{ }}CH(C{H_3}){\rm{ }} - {\rm{ }}C{H_3} =  > {\rm{ }}2 - Metyl{\rm{ }}Propan\)

2,

\(\begin{array}{l}
{n_{ankin}} = 0,05mol\\
{n_{{O_2}}} = 0,125mol\\
{C_n}{H_{2n - 2}} + \dfrac{{3n - 1}}{2}{O_2} \to nC{O_2} + (n - 1){H_2}O\\
{n_{{O_2}}} = \dfrac{{3n - 1}}{{40}} = 0,125 \to n = 2\\
{C_2}{H_2}
\end{array}\)

\(CH \equiv CH:etin\)

3,

\(\begin{array}{l}
{n_{anken}} = 0,05mol\\
{n_{C{O_2}}} = 0,25mol\\
C = \dfrac{{{n_{C{O_2}}}}}{{{n_{anken}}}} = 5\\
{C_5}{H_{10}}\\
{C_5}{H_{10}} + \dfrac{{15}}{2}{O_2} \to 5C{O_2} + 5{H_2}O\\
{n_{{O_2}}} = \dfrac{{15}}{2}{n_{anken}} = 0,375mol \to {V_{{O_2}}} = 8,4l\\
{C_5}{H_{10}} + B{r_2} \to {C_5}{H_{10}}B{r_2}\\
{m_{{C_5}{H_{10}}}} = 0,05 \times 70 = 3,5g
\end{array}\)

Khối lượng bình brom sẽ tăng vì anken sẽ tác dụng với brom.Khối lượng tăng 3,5g