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Topic: Gas Law Help  (Read 2141 times)

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Offline pfb250

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Gas Law Help
« on: April 18, 2013, 07:46:42 PM »
Okay, I don't need your help on this entire Lab, that would be ridiculous. I'm just stuck on a certain part, I have no idea what to do. The basic idea of the lab is that we are trying to simulate an airbag by filling a ziploc bag with CO2 (using Na2CO3 and HC2H3O2). The tricky part is that our grade is dependent on ow close we fill the bag with gas without it exploding. Oh, and the teacher decided this would be a great time to start a new policy: we make the procedure of the lab up on our own, every question asked to the teacher about the lab is 5 points off, and if the bag explodes you get a 0.
Now for what I've got so far:
Balanced Equation: 2*Na2CO3 (s) + 4*HC2H3O2 (l) -> 2*CO2 (g) + 2*H2O(l) + 4*NaC2H3O2 (aq)
I then filled the bag with water, the total volume came out to be 740 mL. I then used the ideal gas law PV=nRT.
PV=nRT
(101.3 KPa)(0.740 L)=x(8.314 J/mol*K)(294 K)
x=0.030668 mol
This is the point where I'm stuck. I don't know what to do from here, and I need to find out how much using Na2CO3 and HC2H3O2 I need to fill the bag completely. I don't need an exact answer, I just need to know how to get one, as I am at a loss.

Offline billnotgatez

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Re: Gas Law Help
« Reply #1 on: April 18, 2013, 08:03:06 PM »
2Na2CO3 (s) + 4HC2H3O2 (l)  :rarrow: 2CO2 (g) + 2H2O (l) + 4NaC2H3O2 (aq)

I just rewrote your formula with subscripts to get a better feel for it.

By the way, since this is real life, are you sure the water on the right side of the equation is going to be liquid or gas?

I have not checked your math yet, but I assume you need the grams for each of the starting reactants.

Offline billnotgatez

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Re: Gas Law Help
« Reply #2 on: April 18, 2013, 08:56:48 PM »
This is being fussy but when I see the L K mol and kPa in the variables I expect the same in the constant.
Maybe you were taught differently but you are fortunate that
8.3144621 L kPa K−1 mol−1
8.3144621 J K−1 mol−1
are the same number

I doubt that it is important but if you use the longer version of 8.3144621 rather than 8.314 you get
~0.0306661

Now your theoretical computation is a good start but we do not want to burst the bag.
So a practical question
Do you think that the bag of Carbon dioxide will be the temperature of ice water at the end of the reaction?

Assuming you finally get the amount of Carbon dioxide you need to inflate the bag you can calculate the rest of the compounds in your equation.

How many points do you lose if you slightly under inflate the bag?

Best of luck

Edit
As an additional thought
The normal range of the Earth's air pressure is from  about 98 kPa to 105 kPa
I have not looked to see if that would make a big factor in pressure difference.
« Last Edit: April 18, 2013, 09:15:24 PM by billnotgatez »

Offline Borek

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Re: Gas Law Help
« Reply #3 on: April 19, 2013, 03:11:54 AM »
Balanced Equation: 2*Na2CO3 (s) + 4*HC2H3O2 (l) -> 2*CO2 (g) + 2*H2O(l) + 4*NaC2H3O2 (aq)

Not bad, but it can be still simplified.

Quote
I then filled the bag with water, the total volume came out to be 740 mL. I then used the ideal gas law PV=nRT.
PV=nRT
(101.3 KPa)(0.740 L)=x(8.314 J/mol*K)(294 K)
x=0.030668 mol
This is the point where I'm stuck. I don't know what to do from here, and I need to find out how much using Na2CO3 and HC2H3O2 I need to fill the bag completely. I don't need an exact answer, I just need to know how to get one, as I am at a loss.

You don't need exact amounts of BOTH - it is enough that one is a limiting reagent.

In what form will the reagents be?

0.0307 moles, or even 0.031 moles is accurate enough - there will be way too many unknown factors (exact temperature, exact pressure, reagent purity) to bother with the less significant digits.
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