Đáp án:
\(\begin{array}{l}
CTPT:{C_3}{H_6}\\
CTCT:\\
C{H_2} = CH - C{H_3}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2{C_x}{H_y} + \dfrac{{4x + y}}{2}{O_2} \xrightarrow{t^0} 2xC{O_2} + y{H_2}O\\
x = \dfrac{{{V_{C{O_2}}}}}{{{V_{{C_x}{H_y}}}}} = \dfrac{{4,5}}{{1,5}} = 3\\
{V_{{C_x}{H_y}}} = (2:\dfrac{{4x + y}}{2}){V_{{O_2}}}\\
\Rightarrow \dfrac{4}{{4 \times 3 + y}} = \dfrac{{1,5}}{{6,75}}\\
\Rightarrow y = \dfrac{{4 \times 6,75 - 1,5 \times 12}}{{1,5}} = 6\\
\Rightarrow CTPT:{C_3}{H_6}\\
CTCT:\\
C{H_2} = CH - C{H_3}
\end{array}\)
