hello!
this is a question from a previous quiz for chem 1a03

What is the volume of the oxygen produced when 1.00L carbon dioxide reacts with 20.0g of Na2O2 based on the following unbalanced reaction at 25C and 2.00 atm?
Na2O2 + CO2 -> O2 + Na2CO3

this is my attempt at it, and the answer i get was present in the options except that it was wrong -_-

here is my attempt. can somebody correct me? (please and thanks

balance the equation:
2Na2O2 + 2CO2 -> O2 + 2Na2CO3

T = 298.15 K
P = 2 atm
R = 0.08206

find moles of O2:
20 g Na2O2 x (1 mol Na2O2/77.9 g Na2O2) x (1 mol O2/ 2 mol Na2O2) = 0.1283 moles O2

use the ideal gas law:

PV = nRT

V = nRT/P = (0.1283 mol)(0.08206)(298.15) / 2 atm = 1.57 L

0.49 L?


This was a lot harder for me than it should have been considering school ended just 3 weeks ago.. -_-

ya thats the right answer. do u mind explaining how u got it?

Originally Posted by RememberTwce

0.49 L?


This was a lot harder for me than it should have been considering school ended just 3 weeks ago.. -_-

You're brave to even attempt it. I don't even want to look at it. xD

Originally Posted by epikness

ya thats the right answer. do u mind explaining how u got it?

You didn't consider that CO2 might be the limiting reagent.

Use PV=nRT with 1.00L of CO2 and you will get the moles of CO2 to be 0.08.

This means that CO2 is the limiting reagent and the mols of O2 will be 0.04.

Then just use PV=nRT:

PV=nRT
(2)(V) = (0.04)(0.08206)(298.1 5)
V = 0.49 L

Originally Posted by Faer

You're brave to even attempt it. I don't even want to look at it. xD

It's sad how unsure I was of myself.

I was refreshing every 5 seconds, waiting for the reply that said "Sorry 0.49 isn't the right answer".

Originally Posted by RememberTwce

It's sad how unsure I was of myself.

I was refreshing every 5 seconds, waiting for the reply that said "Sorry 0.49 isn't the right answer".

Luckily you were exempted from that li'l horror. xD

This is a funny question... gases and water are almost never limiting...

But I got 0.49 L too, if any cares, lol.