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24ml氧气和氢气混合气点燃充分反应后.再恢复到原来状态.剩余气体3ml.则原来混合气体中氢气和氧气各为多少.
人气:454 ℃ 时间:2020-06-17 14:00:05
解答
2H2 + O2 =点燃= 2H2O (l)
24-3 =21 ml;参与反应的气体体积为:V(H2) = 21*2/(2+1) = 14ml; V(O2) = 21*1/(2+1) = 7 ml.
有两种情况:
1.H2过剩,剩余3ml为H2:V(H2) = 17 ml; V(O2) = 7 ml.
2.O2 过剩,剩余3ml为O2:V(H2)=14ml; V(O2) = 10ml.
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