Question

1mol/LCH4(g)完全燃烧生成CO2和液态水释放出的热量为802kj,但当不完全燃烧生成CO2和液态水时,放出的热

Answer 1

CH4+nO2=xCO2+(1-x)CO+2H2O;731.25

2x+(1-x)+2=2n,得x=2n-3

CH4+nO2=（2n-3）CO2+(4-2n)CO+2H2O;731.25

CH4+2O2=CO2+2H2O;802

CH4+(3/2)O2=CO+2H2O;519

用十字交叉法.

不清楚了,应该可以这样做~~

CO2 : 802 212

731 =3:1

CO : 519 71

x:(1-x)=3:1则x=0.75,得n=1.875