CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

A sample of pure CuO was reduced with H2 gas and H2O formed was collected in a 44.8 L flask containing dry N2. At 27oC, the total pressure containing N2 and H2O was 1.0 atm. The relative humidity in the flask was 80%. The vapour pressure of water at 27o C is 25 mm.
How many grams of CuO was reduced? (as nearest integer)

Open in App
Solution

The reaction is as follows:
CuO+H2Cu+H2O(g)
Volume of dry N2=44.8 L at 27oC
Total pressure (N2+H2O)=1.0 atm
Vapour pressure of H2O=25 mm
Humidity =80%
Pressure due to H2O vapour=0.8×25760 mm
Moles of H2O vapour=PVRT=0.8×25760×44.80.082×300 =0.0478
According to the above equation,
1 mol of H2O is obtained from 1 mol of CuO.
Therefore, moles of CuO reduced=0.0478 mol
Weight of CuO=0.0478×(63.5+16)=3.8 g
So, the nearest integer value is 4 g.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Van't Hoff Factor
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon