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Nikit Saraf Grade: 11
        

 


A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the equation, CuO + H2 Ò Cu + H20 ; Oxygen then oxidizes the copper formed : Cu + ½O2 Ò CuO. 100 cm³ of the mixture measured at 25°C and 750 mm yields 84.5 cm³of dry oxygen measured at 25°C and 750 mm after passing over CuO and drying agent. What is the mole present of H2 in the mixture?

7 years ago

Answers : (1)

askiitian.expert- chandra sekhar
10 Points
										

Hi nikit,


V=100 ml


let H2 is x ml


then xml of H2 and x/2 ml O2 are consumed.....


then (100-x) ml O2 initially


and 100-x-x/2 ml O2 finally


therefore,


100-x-x/2 = 84.5


3x/2 = 15.5


x=31/3


mole% of H2 in the mixture is (31/3)*100/100 = 10.33%

7 years ago
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