Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{C{H_4}}} = 25\% \\
\% {V_{{C_2}{H_2}}} = 75\% \\
b)\\
{V_{{O_2}}} = 2,66l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{M_{hh}} = 11,75 \times 2 = 23,5g\\
{n_{hh}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{m_{hh}} = 0,4 \times 23,5 = 9,4g\\
hh:C{H_4}(a\,mol),{C_2}{H_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,4\\
16a + 26b = 9,4
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,4\\
\% {V_{C{H_4}}} = \dfrac{{0,1 \times 22,4}}{{8,96}} \times 100\% = 25\% \\
\% {V_{{C_2}{H_2}}} = 100 - 25 = 75\% \\
b)\\
{n_{hh}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
\Rightarrow {n_{C{H_4}}} = 0,0125\,mol;{n_{{C_2}{H_2}}} = 0,0375\,mol\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 0,0125 \times 2 + 0,0375 \times \dfrac{5}{2} = 0,11875\,mol\\
{V_{{O_2}}} = 0,11875 \times 22,4 = 2,66l
\end{array}\)
