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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 + H 2 0 ; Oxygen then oxidizes the copper formed : Cu + ½O 2 Ò CuO. 100 cm ³ of the mixture measured at 25 ° C and 750 mm yields 84.5 cm ³ of dry oxygen measured at 25 ° C and 750 mm after passing over CuO and drying agent. What is the mole present of H 2 in the mixture?

11 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%

11 years ago
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