$\overline{M}_A$= 3,75.2= 7,5
Gọi x là mol $H_2$, y là mol CO
Ta có: $\frac{2x+ 28y}{x+y}$= 7,5
<=> 5,5x= 20,5y
<=> $\frac{x}{y}$= $\frac{20,5}{5,5}$= $\frac{41}{11}$
Vậy nếu mol $H_2$ là 41x thì mol CO là 11x
H2+ $\frac{1}{2}$O2 (t*)-> H2O
CO+ $\frac{1}{2}$O2 (t*)-> CO2
$n_{O_2}$= 0,25 mol
=> 20,5x+ 5,5x= 0,25
<=> x= $\frac{1}{104}$
=> $n_{H_2}$= $\frac{41}{104}$ mol; $n_{CO}$= $\frac{11}{104}$ mol
%H2= (41/104 .100)/(41/104 + 11/104 )= 78,8%
%CO= 21,2%