Đáp án:
\(\begin{array}{l}
a)\\
CTPT:{C_3}{H_4}\\
b)\\
{V_{{O_2}}} = 8,96l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_n}{H_{2n - 2}} + \dfrac{{3n - 1}}{2}{O_2} \to nC{O_2} + (n - 1){H_2}O\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2mol\\
{n_{Akin}} = {n_{C{O_2}}} - {n_{{H_2}O}} = 0,3 - 0,2 = 0,1mol\\
n = \frac{{{n_{C{O_2}}}}}{{{n_{Akin}}}} = \frac{{0,3}}{{0,1}} = 3\\
\Rightarrow CTPT:{C_3}{H_4}\\
b)\\
{C_3}{H_4} + 4{O_2} \xrightarrow{t^0} 3C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 4{n_{Akin}} = 0,4mol\\
{V_{{O_2}}} = 0,4 \times 22,4 = 8,96l
\end{array}\)