Đáp án:

\(\begin{array}{l}
\% {V_{{C_3}{H_4}}} = 83,33\% \\
\% {V_{{C_4}{H_6}}} = 16,67\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{n_{C{O_2}}} = \dfrac{{21,28}}{{22,4}} = 0,95\,mol\\
{n_{{H_2}O}} = \dfrac{{11,7}}{{18}} = 0,65\,mol\\
{n_{C{O_2}}} > {n_{{H_2}O}} \Rightarrow CTTQ\,A:{C_n}{H_{2n - 2}}\\
{n_A} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
n = \dfrac{{0,95}}{{0,3}} = 3,167 \Rightarrow hh:{C_3}{H_4}(a\,mol),{C_4}{H_6}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
3a + 4b = 0,95
\end{array} \right.\\
 \Rightarrow a = 0,25;b = 0,05\\
\% {V_{{C_3}{H_4}}} = \dfrac{{0,25}}{{0,3}} \times 100\%  = 83,33\% \\
\% {V_{{C_4}{H_6}}} = 100 - 83,33 = 16,67\% 
\end{array}\)