Đốt cháy hoàn toàn 70,1 gam hỗn hợp X gồm C4H6O, C5H6O2, C3H6O3, C4H8O4 cần dùng 68,32 lít O2 (đktc).

* Hướng dẫn giải

\(\left. \begin{array}{l}
{C_3}{H_6}{O_3} \leftrightarrow 3C{H_2}O\\
{C_4}{H_8}{O_4} \leftrightarrow 4C{H_2}O
\end{array} \right\} \Rightarrow \)  Qui đổi C₃H₆O₃ và C₄H₈O₄ thành CH₂O

\(\begin{array}{l}
{C_4}{H_6}O \leftrightarrow C{H_2}O + {C_3}{H_4}\\
{C_5}{H_6}{O_2} \leftrightarrow 2C{H_2}O + {C_3}{H_2}
\end{array}\) 

→ Hỗn hợp X gồm C₃H₄, C₃H₂ và CH₂O

Số mol O₂ là: \({n_{{O_2}}} = \frac{{68,32}}{{22,4}} = 3,05{\rm{ mol}}\) 

\(\begin{array}{l}
 \to {m_X} + {m_{{O_2}}} = {m_{C{O_2}}} + {m_{{H_2}O}} \Rightarrow 70,1 + 32.3,05 = {m_{C{O_2}}} + {m_{{H_2}O}}\\
 \Rightarrow {m_{C{O_2}}} + {m_{{H_2}O}} = 167,7{\rm{ gam}}
\end{array}\) 

\({m_{dd{\rm{ giam}}}} = {m_{BaC{O_3} \downarrow }} - ({m_{C{O_2}}} + {m_{{H_2}O}}) \Rightarrow 393,75 = {m_{BaC{O_3} \downarrow }} - 167,7 \Rightarrow {m_{BaC{O_3} \downarrow }} = 561,45{\rm{ gam}}\) 

\( \Rightarrow {n_{BaC{O_3} \downarrow }} = \frac{{561,45}}{{197}} = 2,85{\rm{ mol}}\)

\({m_{C{O_2}}} + {m_{{H_2}O}} = 167,7 \Rightarrow 44.2,85 + 18.{n_{{H_2}O}} = 167,7 \Rightarrow {n_{{H_2}O}} = 2,35{\rm{ mol}}\) 

Sơ đồ phản ứng:

\(\underbrace {\left\{ \begin{array}{l}

\underbrace {{C_3}{H_4}}_{a{\rm{ mol}}}\\
\underbrace {{C_3}{H_2}}_{b{\rm{ mol}}}\\
C{H_2}O
\end{array} \right\}}_{70,1{\rm{ gam X}}} + \underbrace {{O_2}}_{3,05{\rm{ mol}}} \to \underbrace {C{O_2}}_{2,85{\rm{ mol}}} + \underbrace {{H_2}O}_{2,35{\rm{ mol}}}\) 

\(\begin{array}{l}
 \to {n_{C{H_2}O}} + 2.{n_{{O_2}}} = 2.{n_{C{O_2}}} + {n_{{H_2}O}} \Rightarrow {n_{C{H_2}O}} + 2.3,05 = 2.2,85 + 2,35\\
 \Rightarrow {n_{C{H_2}O}} = 1,95{\rm{ mol}}
\end{array}\) 

\( \to 3.{n_{{C_3}{H_4}}} + 3.{n_{{C_3}{H_2}}} + 2.{n_{C{H_2}O}} = {n_{C{O_2}}} \Rightarrow 3a + 3b + 1,95 = 2,85{\rm{ (I)}}\) 

\(\begin{array}{l}
 \to 4.{n_{{C_3}{H_4}}} + 3.{n_{{C_3}{H_2}}} + 2.{n_{C{H_2}O}} = 2.{n_{{H_2}O}} \Rightarrow 4a + 2b + 2.1,95 = 2.2,35{\rm{ (II)}}\\
 \to a = 0,1{\rm{ mol; b  =  0,2 mol}}
\end{array}\) 

\(\begin{array}{l}
{n_{{C_4}{H_6}O}} = {n_{{C_3}{H_4}}} = 0,1{\rm{ mol}}\\
{\rm{\% }}{{\rm{m}}_{{C_4}{H_6}O}} = \frac{{{m_{{C_4}{H_6}O}}}}{{{m_X}}}.100 = \frac{{70.01}}{{70,1}}.100 = 9,99\%  \approx 10\% 
\end{array}\)