\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT: CH4 + 2O2 \(\underrightarrow{t^o}\) CO2 + 2H2O
Trước 0,1 0,15 0 0 mol
Trong 0,075 0,15 0,075 0,15 mol
Sau 0,025 0 0,075 0,15 mol
\(V_{CO_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\)
\(V_{H_2O\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)