Kanker:
I don’t know where you got this problem, but from its flavor I have to assume you got it out of a text book as a school homework problem. That makes you a student – and one that has not been paying that much attention in class. Or, perhaps, you have a rotten prof who doesn’t take the time to teach his chemical engineering students how to dissect and operate on a simple 1st year engineering problem. If the latter is the case, I would raise holy hell about the amount of money being paid for an education that doesn’t; teach even the basics that will be sorely needed in the later, tougher courses to come.
I refuse to do homework for students. It’s a slur against the student and damages his/her chances of ever learning how to solve engineering problems on his/her own – a cursed malady that will surely lead to his/her flunking out of engineering or never being able to carry out their career successfully. However, what I can do is show you how simple and direct the solution is – if you take the time and study the methodology and basic principles applied. Hopefully, this will help you (& other students) out in future similar problems.
All experienced chemical engineers know one, universal, and often applied rule: when you are stuck on a process problem, always stop and make a heat and material balance. In your case, there is no reaction and there is no heat transfer. The exercise immediately reduces itself to a simple, high school equation:
Mols of CO2 entering = Mols of CO2 exiting
(Mols CO2 in Stream 1) + (Mols CO2 in Stream 2) = (Mols CO2 in Stream 3)
(0.012) (Mols Stream 1) + (Mols Stream 2) = (0.034) (Mols Stream 3) = (0.034) (Mols Stream 1 + Mols Stream 2)
You know (or should know) the value of Stream 2.
You now have an equation with only one unknown: Mols Stream 1 (the answer)
I hope this helps you out with similar challenges in the future.