Đáp án:

 V1=17,92 lít; V= 145,6 lít

Giải thích các bước giải:

 a) 

$\begin{gathered}
  {C_4}{H_{10}} + \dfrac{{13}}{2}{O_2} \to 4C{O_2} + 5{H_2}O \hfill \\
  {C_4}{H_8} + 6{O_2} \to 4C{O_2} + 4{H_2}O \hfill \\
  {C_4}{H_6} + \dfrac{{11}}{2}6{O_2} \to 4C{O_2} + 3{H_2}O \hfill \\
  2{H_2} + {O_2} \to 2{H_2}O \hfill \\ 
\end{gathered} $

b) 

$\begin{gathered}
  {M_Y} = 0,8.29 = 23,2 \hfill \\
   \Rightarrow {m_Y} = 23,2.0,5 = 11,6(g) \hfill \\
  {n_{{H_2}O}} = \dfrac{{18}}{{18}} = 1mol \Rightarrow {n_H} = 2{n_{{H_2}O}} = 2mol \hfill \\
   \Rightarrow {m_C} = {m_Y} - {m_Y} = 11,6 - 2 = 9,6(g) \hfill \\
   \Rightarrow {n_{C{O_2}}} = {n_C} = \dfrac{{9,6}}{{12}} = 0,8mol \hfill \\
   \Rightarrow {V_1}={V_{C{O_2}}} = 0,8.22,4 = 17,92(l) \hfill \\ 
\end{gathered} $

Bảo toàn nguyên tố $O$: 

$\begin{gathered}
  2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} = 2.0,8 + 1 = 2,6mol \hfill \\
   \Rightarrow {n_{{O_2}}} = 1,3mol \hfill \\
   \Rightarrow V = {V_{kk}} = 5{V_{{O_2}}} = 5.1,3.22,4 = 145,6\left( l \right) \hfill \\ 
\end{gathered} $