Posted by **Samuel** on Friday, October 15, 2010 at 4:55pm.

If it takes three "breaths" to blow up a balloon to 1.2 L, and each breath supplies the balloon with 0.060 moles of exhaled air, how many moles of air are in a 3.0 L balloon?

i am trying to use v1/n1=v2/n2 but i am stuck, can u help? thank you

- Chemistry Help -
**DrBob222**, Friday, October 15, 2010 at 5:12pm
Will something like this work?

# breaths to blow 3.0 L balloon = 3 breaths x (3.0 L/1.2 L) = 7.5 breaths [or you can use the dimensional analysis method which will be 3.0

L x (3 breaths/1.2 L) =7.5 breaths.

Each breath has 0.06 moles; therefore,

0.060 moles/breath x 7.5 breaths = ?? moles

- Chemistry Help -
**Samuel**, Friday, October 15, 2010 at 5:48pm
using your method i got 0.45moles,is that correct?

- Chemistry Help -
**DrBob222**, Friday, October 15, 2010 at 6:06pm
That was my answer, also. Check my thinking on the problem.

- Chemistry Help -
**Amy**, Friday, October 15, 2010 at 6:11pm
its saying that i am wrong, im stuck

- Chemistry Help -
**Amy**, Friday, October 15, 2010 at 6:37pm
hello?

- Chemistry Help -
**DrBob222**, Friday, October 15, 2010 at 9:38pm
After re-reading the problem very carefully, I'm sticking with my answer. By the way, your v1/n1 = v2/n2 is exactly the same.

1.2/0.18 = 3.0/n2

n2 = 3.0 x (0.18/1.2) = 0.45

I think the problem with your answer is that the number 0.06 moles has only one significant figures in it; therefore, the answer must be rounded to one s.f. So the answer must be rounded from 0.45 moles to either 0.4 or 0.5 depending upon how your teacher has told you to round number when they end in 5. The rule I follow is to "round to the even number"; therefore, I would round to 0.4 but not everyone follows that rule. Let me know if this does or does not take care of the problem. As an interesting aside, note that if we round at the end, I would round to 0.4 BUT if I rounded at the 7.5 mark, I would round that to 8 and 8 x 0.06 = 0.48 which of course rounds to 0.5

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