Em tham khảo nha :
\(\begin{array}{l}
a)\\
CTHH\,X:{C_x}{H_y}{O_z}\\
{n_{C{O_2}}} = \dfrac{m}{M} = \dfrac{{8,8}}{{44}} = 0,2mol\\
{n_{{H_2}O}} = \dfrac{m}{M} = \dfrac{{5,4}}{{18}} = 0,3mol\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_x} + {m_{{O_2}}} = {m_{{H_2}O}} + {m_{C{O_2}}}\\
\Rightarrow {m_{{O_2}}} = 8,8 + 5,4 - 4,6 = 9,6g\\
{n_{{O_2}}} = \dfrac{m}{M} = \dfrac{{9,6}}{{32}} = 0,3mol\\
{n_C} = {n_{C{O_2}}} = 0,2mol\\
{n_H} = 2{n_{{H_2}O}} = 0,6mol\\
{n_O} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,1mol\\
{M_X} = 23{M_{{H_2}}} = 46dvC\\
{n_X} = \dfrac{m}{M} = \dfrac{{4,6}}{{46}} = 0,1mol\\
x = \dfrac{{{n_C}}}{{{n_X}}} = \dfrac{{0,2}}{{0,1}} = 2\\
y = \dfrac{{{n_H}}}{{{n_X}}} = \dfrac{{0,6}}{{0,1}} = 6\\
z = \dfrac{{{n_O}}}{{{n_X}}} = \dfrac{{0,1}}{{0,1}} = 1\\
\Rightarrow CTHH:{C_2}{H_6}O\\
b)\\
{C_2}{H_6}O + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 3{H_2}O\\
c)\\
\% C = \dfrac{{2{M_C}}}{{{M_X}}} \times 100\% = \dfrac{{2 \times 12}}{{46}} \times 100\% = 52,17\% \\
\% H = \dfrac{{6{M_H}}}{{{M_X}}} \times 100\% = \dfrac{{6 \times 1}}{{46}} \times 100\% = 13,04\% \\
\% O = \dfrac{{{M_O}}}{{{M_X}}} \times 100\% = \dfrac{{16}}{{46}} \times 100\% = 34,79\%
\end{array}\)
