A flask of 1.5 Lcapacity 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?

A flask of 1.5 Lcapacity contains 400 mg of O_(2) and 60 mg of H_(2) a

1.	A flask of 1.5 Lcapacity 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?
Answer» <P> Solution :`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`