Đáp án:

a) $ \% {V_{{C_2}{H_4}}} = \% {V_{{H_2}}} = 50\% $

 b) $\% {V_{{C_2}{H_4}}} = \% {V_{{H_2}}} = 40\% ;\% {V_{{C_2}{H_6}}} = 20\% $

c) 33,33%

Giải thích các bước giải:

 a) ${M_X} = 7,5.2 = 15$

$\begin{array}{*{20}{c}}
  {{C_2}{H_4}:28}&{}&{}&{}&{13} \\ 
  {}& \searrow &{}& \nearrow &{} \\ 
  {}&{}&{15}&{}&{} \\ 
  {}& \nearrow &{}& \searrow &{} \\ 
  {{H_2}:2}&{}&{}&{}&{13} 
\end{array} \Rightarrow \dfrac{{{n_{{C_2}{H_4}}}}}{{{n_{{H_2}}}}} = \dfrac{1}{1}$

$ \Rightarrow \% {V_{{C_2}{H_4}}} = \% {V_{{H_2}}} = 50\% $

b) 

PTHH: ${C_2}{H_4} + {H_2} \to {C_2}{H_6}$

Bđ:      1       1

Pư:      x        x        x

Sau:   1-x    1-x      x

$ \Rightarrow {n_Y} = 2 - x$

Bảo toàn khối lượng: ${m_X} = {m_Y} \Leftrightarrow {n_X}.{M_X} = {n_Y}.{M_Y}$

$ \Rightarrow {n_Y} = \dfrac{{2.15}}{{18}} = \dfrac{5}{3} \Rightarrow 2 - x = \dfrac{5}{3} \Rightarrow x = \dfrac{1}{3}$

$\begin{gathered}
   \Rightarrow {n_{{C_2}{H_4}}} = \dfrac{2}{3};{n_{{H_2}}} = \dfrac{2}{3};{n_{{C_2}{H_6}}} = \dfrac{1}{3} \hfill \\
   \Rightarrow \% {V_{{C_2}{H_4}}} = \% {V_{{H_2}}} = \dfrac{2}{3}:\dfrac{5}{3}.100\%  = 40\%  \hfill \\
  \% {V_{{C_2}{H_6}}} = 20\%  \hfill \\ 
\end{gathered} $

c) $H = \dfrac{{{n_{{H_2}(pu)}}}}{{{n_{{H_2}(bd)}}}}.100\%  = \dfrac{x}{1}.100\%  = 33,33\% $