Em tham khảo nha :

\(\begin{array}{l}
a)\\
{n_{{C_2}{H_6}}}:{n_{{O_2}}} = 1:2\\
{M_{hh}} = \dfrac{{30 \times 1 + 32 \times 2}}{3} = \dfrac{{94}}{3}dvC\\
{d_{hh/{H_2}}} = \dfrac{{94}}{3}:2 = \dfrac{{47}}{3}\\
b)\\
hh:{C_2}{H_6}(a\,mol),{O_2}(b\,mol)\\
\left\{ \begin{array}{l}
2a - b = 0\\
30a + 32b = 112,8
\end{array} \right.\\
 \Rightarrow a = 1,2;b = 2,4\\
{m_{{C_2}{H_6}}} = 1,2 \times 30 = 36g\\
{m_{{O_2}}} = 2,4 \times 32 = 76,8g
\end{array}\)

C3

a. Lấy 1 mol hhX

=>$n_{O2}$=$\frac{2}{3}$ mol

=>$n_{C2H6}$=$\frac{1}{3}$ mol

$M_{X}$= $\frac{mC2H6 + mO2}{nC2H6+nO2}$=$\frac{1/3.30+2/3.32}{1}$=$\frac{94}{3}$

=>$\frac{M_{X}}{H2}$=$\frac{94}{3}$:2=$\frac{47}{3}$

b. Gọi a là số mol C2H6 => $n_{O2}$ là 2a mol

Ta có: 30a + 32.2a=112,8

          => a=1,2 

=>$\left \{ {{mC2H6=36g} \atop {mO2=76,8g}} \right.$