Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{163,68}}{{44}} = 3,72mol\\
{n_{{O_2}}} = a\,mol\\
BT\,Oxi:\\
2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}\\
\left\{ \begin{array}{l}
2{n_{{O_2}}} - {n_{{H_2}O}} = 7,44\\
32{n_{{O_2}}} - 18{n_{{H_2}O}} = 107,88
\end{array} \right.\\
\Rightarrow {n_{{O_2}}} = 6,51mol;{n_{{H_2}O}} = 5,58mol\\
{n_R} = {n_{{H_2}O}} - {n_{C{O_2}}} = 1,86mol\\
CTPT\,R:{C_n}{H_{2n + 2}}\\
CTPT\,Q:{C_m}{H_{2m}}\\
{n_{C{O_2}}} = n \times {n_R} + m \times {n_Q} = 3,72\\
1,86n + m \times {n_Q} = 3,72\\
\Rightarrow n < 2 \Rightarrow n = 1 \Rightarrow CTPT\,R:C{H_4}\\
b)\\
{n_Q} = \dfrac{{3,72 - 1,86}}{m} = \dfrac{{1,86}}{m}\\
{m_Q} = 55,8 - 1,86 \times 16 = 26,04g\\
{M_Q} = \dfrac{{{m_Q}}}{{{n_Q}}} = 14m\\
{n_{C{O_2}(Q)}} = 3,72 - 1,86 \times 1 = 1,86mol
\end{array}\)
Câu b bị thiếu dữ kiện không giải được
