Đáp án:
\(\begin{array}{l}
\% {V_{C{H_4}}} = 33,33\% \\
\% {V_{{H_2}}} = 66,67\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{20}}{{100}} = 0,2\,mol\\
{n_{C{H_4}}} = {n_{C{O_2}}} = 0,2\,mol\\
\% {V_{C{H_4}}} = \dfrac{{0,2 \times 22,4}}{{13,44}} \times 100\% = 33,33\% \\
\% {V_{{H_2}}} = 100 - 33,33 = 66,67\%
\end{array}\)
