\(n_{CH4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(V_{O2}=28.20\%=5,6\)
\(\Rightarrow n_{O2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(PTHH:CH_4+2O_2\rightarrow CO_2+2H_2O\)
Trước_____0,1__ 0,25______________
Phứng__0,1___0,2________________
Sau______ 0 ____ 0,05 ___ 0,1 __ 0,2
\(\Rightarrow\) O2 dư, nO2 dư=0,05mol