Đáp án:
\(\% {m_{{C_4}{H_{10}}}} = 61,7\% \)
Giải thích các bước giải:
\(\left\{ \begin{gathered}
{C_4}{H_{10}}:a\ mol \hfill \\
{C_5}{H_{12}}:b\ mol \hfill \\
\end{gathered} \right.\)
$\text{Đốt cháy khí gas:}$
\({C_4}{H_{10}}\xrightarrow{{{t^o}}}4C{O_2} + 5{H_2}O\\\hspace{0,5cm}a→\hspace{0,7cm}4a\hspace{1,4cm}5a\)
\({C_5}{H_{12}}\xrightarrow{{{t^o}}}5C{O_2} + 6{H_2}O\\\hspace{0,5cm}b→\hspace{0,7cm}5b\hspace{1,4cm}6b\)
$\text{Theo đề bài ta có:}$
\(\dfrac{{{n_{C{O_2}}}}}{{{n_{{H_2}O}}}} = \dfrac{{4a + 5b}}{{5a + 6b}} = \dfrac{{13}}{{16}} \to \dfrac{a}{b} = \dfrac{2}{1}\)
\(\% {m_{{C_4}{H_{10}}}} = \dfrac{{58a}}{{58a + 72b}}.100 = \dfrac{{58a}}{{58a + 36a}}.100 = 61,7\% \)
