012109_lecture - Gas Rxn Stoichiometry Effusion/Diffusion...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
January 21 st 2009 , 2009 •Gas Rxn Stoichiometry •Effusion/Diffusion •Real Gases
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CO 2.0 L H 2 0.50 L 3.0 atm 2.0 atm CO (g) + 2 H 2 (g) -> CH 3 OH (g) at 0C What is the pressure of CH 3 OH produced? What is the total pressure in the vessel after the reaction?
Image of page 2
CO (g) + 2 H 2 (g) -> CH 3 OH (g) at 0C Pressure of CH 3 OH after reaction: Mols CO and H 2 -> limiting reactant -> mols CH 3 OH -> pressure CH 3 OH PV RT PV/RT PV=nRT, n=PV/RT n CO =(3.0 atm*2.0 L)/(0.08206 L atm/mol K * 273 K)=.268 mol This amount of CO would produce 0.268 moles CH3OH ( * )/( * ) l n H2 =(2.0 atm*0.50 L)/(0.08206 L atm/mol K * 273 K)=.0446 mol This amount of H 2 would produce 0.0223 moles CH 3 OH-Liming reactant P CH3OH =nRT/V=(0.0223 moles * 0.08206 L atm/mol K * 273 K)/ 2.5 L=0.20 atm Total pressure after reaction: CO was the non-limiting reactant some will be leftover CO was the non-limiting reactant, some will be leftover Producing 0.0223 mol CH 3 OH will require 0.0223 mol CO 0.268 mol CO – 0.0223 mol used = 0.246 mol leftover P RT/V 2 20 P=nRT/V=2.20 atm Total pressure=P CH3OH + P CO =0.20 atm + 2.20 atm= 2.40 atm
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern