a) CH4+O2suy ra CO2+H2O
b) CH4+2O2suy ra CO2+2H2O
c) MCH4 = 16
% mC = \(\dfrac{12}{16}.100\%=75\%\)
% mH = 25%
a) CH4 + O2 ---➢ CO2 + H2O
b) CH4 + 2O2 \(\underrightarrow{to}\) CO2 + 2H2O
c) \(M_{CH_4}=12+4=16\left(g\right)\)
\(\%C=\dfrac{12}{16}\times100\%=75\%\)
\(\%O=\dfrac{4}{16}\times100\%=25\%\)