Cách 1: nH2O = 2a (mol), nCO2 = a (mol)
Hydrocarbon là alkan do nH2O > nCO2
⇒ nCnH2n+2 = nH2O - nCO2 = a (mol)
\(\Rightarrow n=\dfrac{n_{CO_2}}{n_{C_nH_{2n+2}}}=1\)
→ CH4
Cách 2:
\(2C_xH_y+\dfrac{4x+y}{2}O_2\underrightarrow{t^o}2xCO_2+yH_2O\)
Ta có: \(\dfrac{n_{CO_2}}{n_{H_2O}}=\dfrac{1}{2}=\dfrac{2x}{y}\Rightarrow\dfrac{x}{y}=\dfrac{1}{4}\)
→ CH4