Đáp án:
\(\begin{array}{l}
1)\\
hh:{C_3}{H_8};{C_4}{H_{10}}\\
2)\\
{V_{{C_3}{H_8}}} = 2,24l\\
{V_{{C_4}{H_{10}}}} = 4,48l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
{n_{C{O_2}}} = \dfrac{{24,64}}{{22,4}} = 1,1\,mol\\
{n_{{H_2}O}} = \dfrac{{25,2}}{{18}} = 1,4\,mol\\
{n_{{H_2}O}} > {n_{C{O_2}}} \Rightarrow \text{ Hỗn hợp ankan } \\
CTTQ:{C_n}{H_{2n + 2}}\\
{n_{hh}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 1,4 - 1,1 = 0,3\,mol\\
{M_{hh}} = \dfrac{{16}}{{0,3}} = 53,33g/mol\\
\Rightarrow 14n + 2 = 53,33 \Leftrightarrow n = 3,67\\
\Rightarrow hh:{C_3}{H_8};{C_4}{H_{10}}\\
2)\\
hh:{C_3}{H_8}(a\,mol),{C_4}{H_{10}}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
44a + 58b = 16
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
{V_{{C_3}{H_8}}} = 0,1 \times 22,4 = 2,24l\\
{V_{{C_4}{H_{10}}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)
