已知2H2(g)+O2(g)=2H2O(l) ΔH=-571.6kJ·mol-1 CH4(g)+2O2(g)=CO2(g)+2H2O(l)ΔH=-890kJ/mol现有H2与CH4的混合气体112L(标况下),使其完全燃烧生成CO2和H2O(l),若实验测得反应放热3695kJ,则原混合气体中H2与CH4的物质
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![已知2H2(g)+O2(g)=2H2O(l) ΔH=-571.6kJ·mol-1 CH4(g)+2O2(g)=CO2(g)+2H2O(l)ΔH=-890kJ/mol现有H2与CH4的混合气体112L(标况下),使其完全燃烧生成CO2和H2O(l),若实验测得反应放热3695kJ,则原混合气体中H2与CH4的物质](/uploads/image/z/2217093-69-3.jpg?t=%E5%B7%B2%E7%9F%A52H2%28g%29%2BO2%28g%29%3D2H2O%28l%29+%CE%94H%3D-571.6kJ%C2%B7mol-1+CH4%28g%29%2B2O2%28g%29%3DCO2%28g%29%2B2H2O%28l%29%CE%94H%3D-890kJ%2Fmol%E7%8E%B0%E6%9C%89H2%E4%B8%8ECH4%E7%9A%84%E6%B7%B7%E5%90%88%E6%B0%94%E4%BD%93112L%EF%BC%88%E6%A0%87%E5%86%B5%E4%B8%8B%EF%BC%89%2C%E4%BD%BF%E5%85%B6%E5%AE%8C%E5%85%A8%E7%87%83%E7%83%A7%E7%94%9F%E6%88%90CO2%E5%92%8CH2O%EF%BC%88l%29%2C%E8%8B%A5%E5%AE%9E%E9%AA%8C%E6%B5%8B%E5%BE%97%E5%8F%8D%E5%BA%94%E6%94%BE%E7%83%AD3695kJ%2C%E5%88%99%E5%8E%9F%E6%B7%B7%E5%90%88%E6%B0%94%E4%BD%93%E4%B8%ADH2%E4%B8%8ECH4%E7%9A%84%E7%89%A9%E8%B4%A8)
已知2H2(g)+O2(g)=2H2O(l) ΔH=-571.6kJ·mol-1 CH4(g)+2O2(g)=CO2(g)+2H2O(l)ΔH=-890kJ/mol现有H2与CH4的混合气体112L(标况下),使其完全燃烧生成CO2和H2O(l),若实验测得反应放热3695kJ,则原混合气体中H2与CH4的物质
已知2H2(g)+O2(g)=2H2O(l) ΔH=-571.6kJ·mol-1 CH4(g)+2O2(g)=CO2(g)+2H2O(l)ΔH=-890kJ/mol
现有H2与CH4的混合气体112L(标况下),使其完全燃烧生成CO2和H2O(l),若实验测得反应放热3695kJ,则原混合气体中H2与CH4的物质的量的比是
已知2H2(g)+O2(g)=2H2O(l) ΔH=-571.6kJ·mol-1 CH4(g)+2O2(g)=CO2(g)+2H2O(l)ΔH=-890kJ/mol现有H2与CH4的混合气体112L(标况下),使其完全燃烧生成CO2和H2O(l),若实验测得反应放热3695kJ,则原混合气体中H2与CH4的物质
因为H2与CH4的混合气体112L(标况下),即5mol,设H2有x mol,CH4有y mol
则有x+y=5mol ①
2H2(g)+O2(g)=2H2O(l) ΔH=-571.6kJ·mol-1
2 571.6
x m
则2/x=571.6/m→m=285.8x ②
CH4(g)+2O2(g)=CO2(g)+2H2O(l)ΔH=-890kJ/mol
1 890
y n
则1/y=890/n→n=890y ③
由题意知m+n=3695kJ
由①②③得x=1.25 y=3.75
则x:y=1:3
共有气体5mol,都化为1mol状态,得
H2:285.8kJ,CH4:890kJ,混合气体:739kJ
(890-739)/(739-285.8)=1:3
即物质的量之比为1:3
太深奥了