Ta có :
\(n_{CH4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{O2}=\frac{28}{22,4}=1,25\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{^{to}}CO_2+2H_2O\)
0,1____0,2______________________(mol)
Nên sau phản ứng O2 dư
\(\Rightarrow n_{O2_{dư}}=1,25-0,2=1,05\left(mol\right)\)