Question

A sample of pure CuO was reduced with `H_2` gas and `H_(2)O` formed was collected in a 44.8 L flask containing dry `N_(2)`. At `27^@C`, the total pressure containing `N_(2)` and `H_(2)O` was 1.0 atm. The relative humidity in the flask was `80%` The vapour pressure of water at `27^@C` is 25 mm. How many grams of CuO was reduced?

Answer 1

`CuO+H_(2)toCu+H_(2)O(g)`
Volume of dry `N_(2)=44.8 L at 27^@C`
Total pressure `(N_(2)+H_(2)O)=1.0atm`
Vapour pressure of `H_(2)O=25mm`
Humidity `-=80%`
Pressure due to `H_(2)O` vapour `=0.8xx(25)/(760)mm`
Moles of `H_(2)O` vapour`=(PV)/(RT)=(0.8xx(25)/(760)xx44.8)/(0.082xx300)`
According to the above equation:
1 " mol of "`H_(2)O` is obtained from 1 " mol of "CuO.
Therefore, moles of CuO reduced`=0.0478`
Weight of `CuO=0.0478xx(63.5+16)=3.8g`