Có V lít X là \(\left\{ \begin{array}{l}
{C_2}{H_2}:\frac{V}{3}\\
{C_4}{H_4}:\frac{{2V}}{3}
\end{array} \right. + 1,5V\,\,{H_2} \to Y\)

\(Y \to \left\{ \begin{array}{l}
{n_Z} = 0,8(mol)\\
{m_Z} = {m_Y} \to \frac{{{n_Z}}}{{{n_Y}}} = \frac{1}{{1,875}} \to {n_Y} = 1,5(mol)
\end{array} \right. \to V = 0,6(mol)\)

Có \(\left\{ \begin{array}{l}
{n_T} = 0,55(mol) \to n = 0,8 - 0,55 = 0,25(mol)\\
n_\pi ^{trong\,\,T} = 0,15.2 = 0,3(mol)
\end{array} \right.\)

Và  \({n_{Ag}} = 0,3 \to 0,25(mol)\left\{ \begin{array}{l}
CH \equiv CH:0,05\\
CH \equiv C - CH = C{H_2}:a(mol)\\
CH \equiv C - C{H_2} - C{H_3}:b(mol)
\end{array} \right.\)

\( \to a + b = 0,2(mol)\)

\( \to 0,05.2 + 3a + 2b + \underbrace {0,3}_T + \underbrace {0,7}_{{H_2}} = \underbrace {0,2.2}_{{C_2}{H_2}} + \underbrace {0,4.3}_{{C_4}{H_4}}\,\,\, \to \left\{ \begin{array}{l}
a = 0,1\\
b = 0,1
\end{array} \right.\)

\(\begin{array}{l}
 \to m = 0,05.240 + 0,1(159 + 161) = 44(gam)\,\,\\
\, \to \% CAg \equiv CAg = 27,27\% 
\end{array}\)