Đáp án:
66,67% và 33,33%
Giải thích các bước giải:
\(\begin{array}{l}
nhh = \frac{{11,2}}{{22,4}} = 0,5\,mol\\
Mhh = 6,2 \times 2 = 12,4\\
M{C_2}{H_4} = 28\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10,4\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \searrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \nearrow \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Mhh = 12,4\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \nearrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \searrow \\
M{H_2} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15,6\\
= > n{C_2}{H_4}:n{H_2} = 2:3\\
= > n{C_2}{H_4} = 0,2\,mol\\
n{H_2} = 0,3\,mol\\
{C_2}{H_4} + {H_2} \to {C_2}{H_6}\\
n{C_2}{H_6} = n{C_2}{H_4} = 0,2\,mol\\
n{H_2}trongX = 0,3 - 0,2 = 0,1\,mol\\
\% V{C_2}{H_6} = 66,67\% \\
\% V{H_2} = 33,33\%
\end{array}\)
