[supehr]

I just wanted someone to go over my answer for this question and lemme know if it's good!

Q: The reaction of potassium superoxide, KO2 is used in life-support systems to replace CO2(g) in expired air with O2(g). The balanced chemical equation for the reaction is:

4KO₂ (s) + 2CO₂ (g)  2K₂CO₃ (s) + 3O₂ (g)

a) How many moles of O₂ are produced by the reaction of 156g CO₂ with excess KO₂ (s)?
b) How many grams of KO₂ are consumed per 100 g CO₂ removed from expired air?
c) How many O₂ molecules are produced per mg of KO₂ consumed?

My answers: (I am 90% sure i did the steps correctly, i tend to make silly errors in my answers so i just wanted to compare my ans with what some of you get)

a) 5.32 mol O₂
b) 323.09 g KO₂
c) 6.35x10¹⁸ molecules of O₂

Thanks!  i hope i got the right answers 

[Borek]

Looks OK to me.

[supehr]

I also dont understand part b in this question (attatched the question). For a i got an answer of 6.31x10-5 mol/L. I dont understand what to do in b .

[sjb]

I also dont understand part b in this question (attatched the question). For a i got an answer of 6.31x10-5 mol/L. I dont understand what to do in b .

OK, What number of moles will produce 6.31 x 10^-5 mol / L for that volume?