Gọi $n_{H_2}$ là x,$n_{CH_4}$ là y
Ta có: $x+y=5$
$2 x + 16 y = 5 , 2$
$ ⇒ x = 0 , 2 ( m o l ) ; y = 0 , 3 ( m o l )$
$⇒$ % $V_{H_2} = 0 , 2.100 / 0 , 5 = 40$ %
$⇒$ % $V_{CH_2} =100$%$-40$%$=60$%
$b,n_{O_2}=0,9(mol)$
Ta có PTHH:
$2H_2+O_2→2H_2O$
$CH_4+2O_2→CO_2+2H_2O$
$n_{O_2cần}=0,1+0,6=0,7(mol)$
$n_{O_2dư}=0,2(mol)$
$V_{O_2dư}=0,2.22,4=4,48(l)$
Ta có: $n_{CO_2}=n_{CH_4}=0,3(mol)$
$V_{CO_2}=0,3.22,4=6,72(l)$
$V=6,72+4,48=11,2(l)$
$n_{O_2dư}=0,2.32=6,4(gam)$
$n_{CO_2}=0,3.44=13,2(gam)$
$m_{y}=6,4+13,2=19,6(gam)$
%$m_{O_2dư}=6,4.100/19,6≈32,65$%
%$m_{CO_2}≈67,35$%