Bài 1.
\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,15 0,3 ( mol )
\(V_{O_2}=0,3.22,4=6,72l\)
Bài 2.
\(n_{C_2H_4}=\dfrac{11,2}{26}=0,4mol\)
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
0,4 1,2 0,8 0,8 ( mol )
\(m_{CO_2}=0,8.44=35,2g\)
\(m_{H_2O}=0,8.18=14,4g\)
\(V_{kk}=1,2.22,4.5=134,4l\)