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» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :`n_(oO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))=(400xx10^(-3))/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)=0.0125` <br/> `n_(H_(2))=(60xx10^(3))/(2)=0.03` <a href="https://interviewquestions.tuteehub.com/tag/ltbr-2804405" style="font-weight:bold;" target="_blank" title="Click to know more about LTBR">LTBR</a>. Here `O_(2)`. Is the limiting reagent and is completely consumed. <br/> Total moles `n=0.0125+0.03=0.0425mol` <br/> total pressure `P=(0.0425xx0.082xx373)/(1.5)=0.867atm` <br/> `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))` <br/> here `O_(2)` is the limiting reagent and is completely consumed. <br/> Now `p_(H_(2)O)=(0.867xx0.025)/(0.0425)=0.509atm`</body></html>