n $n_{kk}$ =$\frac{89.6}{22.4}$ =4mol->$n_{O2}$ =$\frac{20*4}{100}$ =0.8 mol
Gọi x,y lll số mol H2,CO
a) 2$H_{2}$+ $O_{2}$ $\frac{t0}{}$>2$H_{2}$O
x 0.5x x
2CO+O2$\frac{to}{}$ >2CO2
y 0.5y y
b) Ta có: $\left \{ {{2x+28y=13.6} \atop {0.5x+0.5y=0.8}} \right.$
<=>$\left \{ {{x=1.2} \atop {y=0.4}} \right.$
->mH2=1.2*2=2.4g
mCO=0.4*28=11.2g
vhh=(1.2+0.4)*22.4=35.84l
c) %H2=$\frac{1.2*22.4*100%}{35.84}$ =75%
%CO=100%-75%=25%
