Đáp án:
\(\begin{array}{l}
a)\\
\% {m_C} = 92,31\% \\
\% {m_H} = 7,69\% \\
b)\\
CTDGN:CH\\
c)\\
CTPT:{C_6}{H_6}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{8,8}}{{44}} = 0,2mol\\
{n_{{H_2}O}} = \dfrac{{1,8}}{{18}} = 0,1\,mol\\
{n_C} = {n_{C{O_2}}} = 0,2mol\\
{n_H} = 2{n_{{H_2}O}} = 0,2\,mol\\
\% {m_C} = \dfrac{{0,2 \times 12}}{{2,6}} \times 100\% = 92,31\% \\
\% {m_H} = 100 - 92,31 = 7,69\% \\
b)\\
{n_C}:{n_H} = 0,2:0,2 = 1:1\\
\Rightarrow CTDGN:CH\\
c)\\
CTPTA:\,{(CH)_n}\\
{M_A} = 39 \times {M_{{H_2}}} = 39 \times 2 = 78g/mol\\
\Rightarrow 13n = 78 \Leftrightarrow n = 6\\
CTPT:{C_6}{H_6}
\end{array}\)
