Em tham khảo nha :
\(\begin{array}{l}
1)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O\\
{n_{BaC{O_3}}} = \dfrac{{55,16}}{{197}} = 0,28mol\\
{n_{C{O_2}}} = {n_{BaC{O_3}}} = 0,28mol\\
{n_{hh}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
hh:C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
a + 2b = 0,28
\end{array} \right.\\
\Rightarrow a = 0,12;b = 0,08\\
\% {V_{C{H_4}}} = \dfrac{{0,12 \times 22,4}}{{4,48}} \times 100\% = 60\% \\
\% {V_{{C_2}{H_4}}} = 100 - 60 = 40\% \\
2)\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_C} = {n_{C{O_2}}} = 0,1mol\\
{n_{{H_2}O}} = \dfrac{{2,7}}{{18}} = 0,15mol\\
{n_H} = 2{n_{{H_2}O}} = 0,3mol\\
BT\,Oxi:\\
{m_O} = 2,3 - 0,1 \times 12 - 0,3 \times 1 = 0,8mol\\
{n_O} = \dfrac{{0,8}}{{16}} = 0,05mol\\
{M_A} = 2,875{M_O} = 46dvC\\
{n_A} = \dfrac{{2,3}}{{46}} = 0,05mol\\
CTPT:{C_x}{H_y}{O_z}\\
x = \dfrac{{{n_C}}}{{{n_A}}} = \dfrac{{0,1}}{{0,05}} = 2\\
y = \dfrac{{{n_H}}}{{{n_A}}} = \dfrac{{0,3}}{{0,05}} = 6\\
z = \dfrac{{{n_O}}}{{{n_A}}} = \dfrac{{0,05}}{{0,05}} = 1\\
\Rightarrow CTPT:{C_2}{H_6}O
\end{array}\)
