Đáp án:
\(CTPT:{C_3}{H_8};{C_4}{H_{10}}\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_{\overline n }}{H_{2\overline n + 2}} + \dfrac{{3\overline n + 1}}{2}{O_2} \to \overline n C{O_2} + \overline n + 1{H_2}O\\
{n_{{O_2}}} = \dfrac{{36,8}}{{32}} = 1,15\,mol\\
{n_{hh}} = 1,15 \times \dfrac{1}{{\dfrac{{3\overline n + 1}}{2}}} = \dfrac{{2,3}}{{3\overline n + 1}}(mol)\\
\Rightarrow \dfrac{{10,2}}{{\dfrac{{2,3}}{{3\overline n + 1}}}} = 14\overline n + 2 \Leftrightarrow 14\overline n + 2 = \dfrac{{102}}{{23}}(3\overline n + 1)\\
\Rightarrow \overline n = 3,5 \Rightarrow CTPT:{C_3}{H_8};{C_4}{H_{10}}
\end{array}\)
