a) \(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Xét tỉ lệ: \(\dfrac{0,25}{1}>\dfrac{0,3}{2}\) => CH4 dư, O2 hết
PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
0,15<--0,3--->0,15
\(m_{CH_4\left(dư\right)}=\left(0,25-0,15\right).16=1,6\left(g\right)\)
b) \(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)