Đáp án:

 a) 20,4g

b) 61,76% và 38,24%

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
hh:{C_2}{H_4}(a\,mol),{C_2}{H_2}(b\,mol)\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{C_2}{H_4} + C{l_2} \to {C_2}{H_4}C{l_2}\\
{C_2}{H_2} + 2C{l_2} \to {C_2}{H_2}C{l_4}\\
{n_{{H_2}O}} = \dfrac{{7,2}}{{18}} = 0,4\,mol\\
{n_{C{l_2}}} = \dfrac{{15,68}}{{22,4}} = 0,7\,mol\\
\left\{ \begin{array}{l}
\dfrac{{2a}}{3} + \dfrac{b}{3} = 0,4\\
\dfrac{{2a}}{3} + \dfrac{{4b}}{3} = 0,7
\end{array} \right.\\
 \Rightarrow a = 0,45;b = 0,3\\
{m_{hh}} = 0,45 \times 28 + 0,3 \times 26 = 20,4g\\
b)\\
\% {m_{{C_2}{H_4}}} = \dfrac{{0,45 \times 28}}{{20,4}} \times 100\%  = 61,76\% \\
\% {m_{{C_2}{H_2}}} = 100 - 61,76 = 38,24\% 
\end{array}\)