炽热的炉膛内有反应C(s)+O2(g)=CO2(g);△H=—392kj、mol往炉膛内通入水蒸气时,有如下反应:C(s)+H2O(g)=CO(g)+H2(g);△H=+131kj、mol,CO(g)+0.5 O2(g)=CO2(g);△H=—282kj、mol,H2(g
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![炽热的炉膛内有反应C(s)+O2(g)=CO2(g);△H=—392kj、mol往炉膛内通入水蒸气时,有如下反应:C(s)+H2O(g)=CO(g)+H2(g);△H=+131kj、mol,CO(g)+0.5 O2(g)=CO2(g);△H=—282kj、mol,H2(g](/uploads/image/z/9871122-66-2.jpg?t=%E7%82%BD%E7%83%AD%E7%9A%84%E7%82%89%E8%86%9B%E5%86%85%E6%9C%89%E5%8F%8D%E5%BA%94C%EF%BC%88s%EF%BC%89%2BO2%EF%BC%88g%EF%BC%89%3DCO2%EF%BC%88g%EF%BC%89%EF%BC%9B%E2%96%B3H%3D%E2%80%94392kj%E3%80%81mol%E5%BE%80%E7%82%89%E8%86%9B%E5%86%85%E9%80%9A%E5%85%A5%E6%B0%B4%E8%92%B8%E6%B0%94%E6%97%B6%2C%E6%9C%89%E5%A6%82%E4%B8%8B%E5%8F%8D%E5%BA%94%EF%BC%9AC%EF%BC%88s%EF%BC%89%2BH2O%EF%BC%88g%EF%BC%89%3DCO%EF%BC%88g%EF%BC%89%2BH2%EF%BC%88g%EF%BC%89%EF%BC%9B%E2%96%B3H%3D%2B131kj%E3%80%81mol%2CCO%EF%BC%88g%EF%BC%89%2B0.5+O2%EF%BC%88g%EF%BC%89%3DCO2%EF%BC%88g%EF%BC%89%EF%BC%9B%E2%96%B3H%3D%E2%80%94282kj%E3%80%81mol%2CH2%EF%BC%88g)
炽热的炉膛内有反应C(s)+O2(g)=CO2(g);△H=—392kj、mol往炉膛内通入水蒸气时,有如下反应:C(s)+H2O(g)=CO(g)+H2(g);△H=+131kj、mol,CO(g)+0.5 O2(g)=CO2(g);△H=—282kj、mol,H2(g
炽热的炉膛内有反应C(s)+O2(g)=CO2(g);△H=—392kj、mol
往炉膛内通入水蒸气时,有如下反应:C(s)+H2O(g)=CO(g)+H2(g
);△H=+131kj、mol,CO(g)+0.5 O2(g)=CO2(g);△H=—282kj、mol,H2(g)+0.5 O2(g)=H2O(g);△H=—241kj、mol,由以上反应推断炽热的炉膛内通入水蒸气时
能否使炉火瞬间更旺
能否节省燃料?
炽热的炉膛内有反应C(s)+O2(g)=CO2(g);△H=—392kj、mol往炉膛内通入水蒸气时,有如下反应:C(s)+H2O(g)=CO(g)+H2(g);△H=+131kj、mol,CO(g)+0.5 O2(g)=CO2(g);△H=—282kj、mol,H2(g
会变旺,但是不会节省燃料,因为氢气与氧气最后生成的还是水,所以用水蒸气其实是为了使煤更充分反应,但是绝对不会节省燃料.
从反应热上看可以,其他的我不知道条件,不敢轻易给你下结论
hui