题目内容
已知下列热化学方程式:Fe2O3(s)+3CO(g)=2Fe(s)+3CO2(g)
H="-24.8" kJ·mol-1
3Fe2O3(s)+CO(g)=2Fe3O4(s)+CO2(g)
H="-47.2" kJ·mol-1
Fe3O4(s)+CO(g)="3FeO" (s)+CO2(g)
H="+640.5" kJ·mol-1
则14g CO气体与足量FeO充分反应得到Fe单质和CO2气体时的反应热为( )
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824134937451324.png)
3Fe2O3(s)+CO(g)=2Fe3O4(s)+CO2(g)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824134937451324.png)
Fe3O4(s)+CO(g)="3FeO" (s)+CO2(g)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824134937451324.png)
则14g CO气体与足量FeO充分反应得到Fe单质和CO2气体时的反应热为( )
A.-218 kJ·mol-1 | B.-109kJ·mol-1 | C.+218 kJ·mol-1 | D.+109 kJ·mol-1 |
B
根据①Fe2O3(s)+3CO(g)=2Fe(s)+3CO2(g) △H=-24.8 kJ/mol
②3Fe2O3(s)+ CO(g)==2Fe3O4(s)+ CO2(g) △H=-47.2 kJ/mol
③Fe3O4(s)+CO(g)==3FeO(s)+CO2(g) △H=+640.5 kJ/mol
由①×3-②-③×2可得热化学方程式为:
CO(g)+FeO(s)=" Fe(s)" + CO2(g) △H=-218.00 kJ/mol,故答案为B
②3Fe2O3(s)+ CO(g)==2Fe3O4(s)+ CO2(g) △H=-47.2 kJ/mol
③Fe3O4(s)+CO(g)==3FeO(s)+CO2(g) △H=+640.5 kJ/mol
由①×3-②-③×2可得热化学方程式为:
CO(g)+FeO(s)=" Fe(s)" + CO2(g) △H=-218.00 kJ/mol,故答案为B
![](http://thumb.zyjl.cn/images/loading.gif)
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