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passing 25 ml of a gaseous mixture of N 2 and NO over heated copper, 20 ml of the gas remained. Calculate the percentage of each in the mixture.
hi nikit, 25 ml of N2 + NO 20 ml remained is N2 NO is absorbed therefore 20% NO and 80% N2 are present.
hi nikit,
25 ml of N2 + NO
20 ml remained is N2
NO is absorbed
therefore 20% NO and 80% N2 are present.
Thank you very much for the answer sir. But the book from which i have taken the question says the answer is 60% of N2 and 40% of NO. And i also think but i m not sure about it that your assumption about only N2 being unreacted when N2 and NO is passed over copper is totally correct. I think there also be some quantity of copper or NO which is left unreacted and which can account for the volume of the remaning gases. Sir, so which of the answers do consider as correct? I m also totally confused about it. So please hep me out!
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