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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?
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%
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