已知CH4(g)+2O2(g)==CO2(g)+2H2O(l); △H=-890kJ/mol 现有CH4和CO的混合气体
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已知CH4(g)+2O2(g)==CO2(g)+2H2O(l); △H=-890kJ/mol 现有CH4和CO的混合气体1.5mol .
已知CH4(g)+2O2(g)==CO2(g)+2H2O(l); △H=-890kJ/mol 现有CH4和CO的混合气体1.5mol ,完全燃烧后生成CO2气体和36gH2O液体,放出1030KJ热量.此条件下CO燃烧的热化学方程式
已知CH4(g)+2O2(g)==CO2(g)+2H2O(l); △H=-890kJ/mol 现有CH4和CO的混合气体1.5mol ,完全燃烧后生成CO2气体和36gH2O液体,放出1030KJ热量.此条件下CO燃烧的热化学方程式
![已知CH4(g)+2O2(g)==CO2(g)+2H2O(l); △H=-890kJ/mol 现有CH4和CO的混合气体](/uploads/image/z/3185779-67-9.jpg?t=%E5%B7%B2%E7%9F%A5CH4%28g%29%2B2O2%28g%29%3D%3DCO2%28g%29%2B2H2O%28l%29%3B+%E2%96%B3H%3D-890kJ%2Fmol+%E7%8E%B0%E6%9C%89CH4%E5%92%8CCO%E7%9A%84%E6%B7%B7%E5%90%88%E6%B0%94%E4%BD%93)
答案:CO(g)+1/2O2(g)==CO2(g); △H=-280kJ/mol
原因:CO燃烧的产物只有CO2,所以36gH2O液体全部是由CH4燃烧生成的,由已知的式子可得CH4为1mol,放出的热量为890kJ,
所以CO为0.5mol,燃烧放出的热量为
1030-890=140kJ,
所以1molCO燃烧放热280kJ,
所以CO的燃烧热方程为:
CO(g)+1/2O2(g)==CO2(g); △H=-280kJ/mol
原因:CO燃烧的产物只有CO2,所以36gH2O液体全部是由CH4燃烧生成的,由已知的式子可得CH4为1mol,放出的热量为890kJ,
所以CO为0.5mol,燃烧放出的热量为
1030-890=140kJ,
所以1molCO燃烧放热280kJ,
所以CO的燃烧热方程为:
CO(g)+1/2O2(g)==CO2(g); △H=-280kJ/mol
已知CH4(g)+2O2(g)==CO2(g)+2H2O(l); △H=-890kJ/mol 现有CH4和CO的混合气体
求详解.已知CH4(g)+2O2(g)=CO2(g)+2H2O(l);ΔH=-890 kJ·mol-1,现有CH4和CO
已知2H2(g)+O2(g)==2H2O(l) △H=-571.6KJ/mol CH4(g)+2O2(g)==CO2(g
已知:2H2(g)+O2(g)=2H2O(l) △H=-571.6KJ/mol CH4(g)+2O2(g)=CO2(g)
已知2H2O(L)=2H2(g)+O2(g) △H=+571.6kJ/mol CH4(g)+2O2(g)=CO2(g)+
已知:CH4(g)+2O2(g)=CO2(g)+2H2O(l)△H=-890.3 KJ*mol-1H2(g)+0.5O2
已知 CH4(g)+2O2(g)===2H2O(l)+CO2(g) ΔH=-890.3kj.mol-1求1gH2和1gC
(1)CH4(g)+2O2(g)=CO2(g)+2H2O(l),△H1=-890.3kj/mol
在一定条件下,CO和CH4燃烧的热化学方程式为:CO(g)+1/2O2=CO2(g);△H=-283KJ/mol
已知:2H2(g)+O2(g)═2H2O(l)△H=-571.6kJ•mol-1;CH4(g)+2O2(g)═CO2(g
已知:CH4(g)+2O2(g)═CO2(g)+2H2O(l);△H=-Q1kJ•mol-1,
已知:CH4(g)+2O2(g)=CO2(g)+2H2O(g) ΔH=-Q1 kJ/mol,2H2(g)+O2(g)==