\(V(\mathrm{C_2H_4})=V(\mathrm{H_2})=2{,}5\ \text{л}\), \(V(\mathrm{C_2H_6})=7{,}5\ \text{л}\).
\[ \chi(\mathrm{C_2H_4})=\frac{2{,}5}{10}\cdot100\%=25\% \]
\[ M_{\text{смеси}}=0{,}25\cdot28+0{,}75\cdot30=29{,}5\ \text{г/моль} \] \[ \rho=\frac{29{,}5}{22{,}4}\approx1{,}3\ \text{г/л} \]
\[ V_{\text{прод.}}=2\cdot10+(2\cdot2{,}5+3\cdot7{,}5)=47{,}5\ \text{л} \] \[ \frac{47{,}5}{10}=4{,}75\approx4{,}8 \]
Ответ: 25%; 1,3 г/л; 4,8.
