Giải bài 3 trang 123 SGK Hóa học 11

$\begin{align} & \underset{x\,mol}{\mathop{C{{H}_{4}}}}\,+2{{O}_{2}}\to \underset{x\,mol}{\mathop{C{{O}_{2}}}}\,+2{{H}_{2}}O \\ & \underset{y\,mol}{\mathop{{{C}_{2}}{{H}_{6}}}}\,+\frac{7}{2}{{O}_{2}}\to \underset{2y}{\mathop{2C{{O}_{2}}}}\,+3{{H}_{2}}O \\ & \Rightarrow \,{{n}_{A}}=\frac{3,36}{22,4}=0,15\,mol \\ & {{n}_{C{{O}_{2}}}}=\frac{4,48}{22,4}=0,2\,mol \\ & \end{align}$

$\begin{aligned} & \Rightarrow \left\{ \begin{aligned} & x+y=0,15 \\ & x+2y=0,2 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x=0,10\,mol \\ & y=0,05\,mol \\ \end{aligned} \right. \\ \end{aligned}$

$\%CH_4=\frac{0,1.100}{0,15}=66,67\%$

$\%C_2H_6=33,33\%$