\(PTHH:CH_4+2O_2\underrightarrow{^{t^o}}CO_2+2H_2O\)
Ta có:
\(V_{CH4}=4,48.90\%=4,032\left(l\right)\)
\(\Rightarrow n_{CH4}=\frac{0,432}{22,4}=0,18\left(mol\right)\)
\(\Rightarrow V_{CO2}=V_{CH4}=0,432\left(l\right)\)
\(n_{H2O}=2n_{CH4}=0,18.2=0,36\left(mol\right)\)
\(\Rightarrow m_{H2O}=0,36.18=6,48\left(g\right)\)