Đáp án:
\(\begin{array}{l}
\% {V_{{H_2}}} = \% {V_{CO}} = 20\% \\
\% {V_{C{H_4}}} = 60\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{M_{hh}} = 7,8 \times 2 = 15,6(g/mol)\\
{n_{hh}} = 1\,mol \Rightarrow {m_{hh}} = 15,6 \times 1 = 15,6g\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}\\
hh:{H_2}(a\,mol),C{H_4}(b\,mol),CO(c\,mol)\\
0,5a + 2b + 0,5c = 1,4\\
2a + 16b + 28c = 15,6\\
a + b + c = 1\\
\Rightarrow a = 0,2;b = 0,6;c = 0,2\\
\% {V_{{H_2}}} = \% {V_{CO}} = \dfrac{{0,2}}{1} \times 100\% = 20\% \\
\% {V_{C{H_4}}} = 100 - 20 - 20 = 60\%
\end{array}\)
