Question

A mixture of oxygen and hydrogen is analysed by passing it over hot copper oxide
and through a drying tube. Hydrogen reduces the CuO according to the equation,

CuO + H2, → Cu + H20;

Oxygen then oxidises the copper formed:
Cu + ½02 → CuO


100 cm of the mixture (measured at 25°C and 750 mm) yields 84.5 cm of dry oxygen (measured at 25°C and 750 mm) after passing over CuO and a drying agent. What
is the mole per cent of H, in the mixture?​

Answer 1

0.1033 is the mole per cent of H2 in the mixture.

Explanation:

• Since we know that the volume of the mixture is V=100 ml  while supposing H2 is x ml
• From this, we conclude xml is H2 and we know that x/2 ml O2 is consumed.
• So (100-x) ml O2 is initially attained then 100-x-x/2 ml O2 is the final product.
• Resulting in 100-x-x/2 = 84.5, 3x/2 = 15.5
• We have x=31/3
• By putting the values of percentages we have (31/3)*100/100 = 10.33%

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