CH4 + 2O2 \(\rightarrow\)CO2 + 2H2O (1)
2CO + O2 \(\rightarrow\)2CO2 (2)
nCO2=\(\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
Đặt nCH4=a
nCO=b
Ta có:
\(\left\{{}\begin{matrix}16a+28b=15\\a+b=0,75\end{matrix}\right.\)
a=0,5;b=0,25
mCH4=0,5.16=8(g)
% CH4 =\(\dfrac{8}{15}.100\%=53,3\%\)
% CO=100-53,3=46,7%
b;
Theo PTHH 1 và 2 ta có:
\(\sum n_{O_2}=0,5.2+0,25.\dfrac{1}{2}=1,125\left(mol\right)\)
VO2=1,125.22,4=25,2(lít)