Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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?

 


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?

Grade:11

1 Answers

askiitian.expert- chandra sekhar
10 Points
11 years ago

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%

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free