\(CH_4 + 2O_2-t^o-> CO_2+2H_2O\)
\(2H_2+O_2-t^o-> 2H_2O\)
Gọi a là nCH4, b là nH2
Ta có: n hỗn hợp = \(\dfrac{22,4}{22,4}=1(mol)\)
\(nO_2 = \dfrac{28}{22,4} =1,25 (mol)\)
Theo PTHH, ta có:\(\left\{\begin{matrix}a+b=1\\2a+0,5b=1,25\end{matrix}\right.\)
\(<=>\) \(\left\{\begin{matrix}a=0,5\\b=0,5\end{matrix}\right.\)
%V CH4 = \(\dfrac{0,5.100}{1} = 50\)%
=> %V H2 = 100% - 50% = 50%