A flask of 1.5 L capacity contains 400 mg of `O_(2)` and 60 mg of `H_(2)` at `100^(@)C`. Calculate the total pressure of the gaseous mixture. If the mixture is permitted to react to form water vapour at `100^(@)C`, what materials will be left and what will be their partial pressures?

Explanation

`n_(oO_(2))=(400xx10^(-3))/(32)=0.0125`
`n_(H_(2))=(60xx10^(3))/(2)=0.03` ltbr. Here `O_(2)`. Is the limiting reagent and is completely consumed.
Total moles `n=0.0125+0.03=0.0425mol`
total pressure `P=(0.0425xx0.082xx373)/(1.5)=0.867atm`
`underset(0.005" mol")underset(0.03" mol")(2H_(2)(g))+underset(-)underset(0.0125" mol")(O_(2)(g))tounderset(0.025" mol...After reaction")underset(-" ...initial")(2H_(2)O(g))`
here `O_(2)` is the limiting reagent and is completely consumed.
Now `p_(H_(2)O)=(0.867xx0.025)/(0.0425)=0.509atm`