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?

A sample of pure CuO was reduced with `H_2` gas and `H_(2)O` formed wa

1.	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» `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`

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