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Topic: question about this aqueous solution problem  (Read 13603 times)

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Offline kevintheman

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question about this aqueous solution problem
« on: June 20, 2007, 10:36:12 PM »
here it is:

Consider the balanced chemical equation: Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l).  How many moles of HCl remain if 0.1 mol of Ca(OH)2 react with 0.4 mol of HCl?
   

i think the answer is 0.3 mol, but im not sure, thanks

also im not so sure about this one, if anyone gets a chance.

A flask contains pure Ne at a pressure of 0.50 atm.  If Ar is added to the flask until there is three times as much Ar as Ne, the total pressure in the flask is:

Offline Borek

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Re: question about this aqueous solution problem
« Reply #1 on: June 21, 2007, 02:45:01 AM »
stoichiometric alculations - pay special attention to the paragraph that explains how to read balanced equations.

2nd question - PV=nRT, this is a mixture of two ideal gases.
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Offline millerst

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Re: question about this aqueous solution problem
« Reply #2 on: June 21, 2007, 09:30:42 PM »
Quote
A flask contains pure Ne at a pressure of 0.50 atm.  If Ar is added to the flask until there is three times as much Ar as Ne, the total pressure in the flask is:

I'm not sure this is possible considering PV = nRT needs at least 4 variables to correctly function.

You know pressure: 0.50 atm X 101.3kPa = 50.65 kPa
                                               1 atm
R being a constant is always equal to 8.3

However you are not given a temperature, number of moles or volume. If someone knows how to figure it out, I'd gladly like to know.

Offline enahs

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Re: question about this aqueous solution problem
« Reply #3 on: June 21, 2007, 11:48:58 PM »

However you are not given a temperature, number of moles or volume. If someone knows how to figure it out, I'd gladly like to know.

Not needed.

PV=nRT
V,R,T are the same in both cases, so if you multiply n times 3, the pressure is just times 3. Regardless of what V,R and T happen to be.

Mathematically you can also do it with algebra.

P1V1 =    P2V2   
nRT    (3*n) RT

V2=V1, you know the original pressure (P1). Volume and temperature and number of moles all cancel out neatly when you do the algebra.




You must remember in either way you do it (and clearly the first way is the easiest) that, that is the pressure of the new system, not the total system.



And you are correct in that R is a constant, but it is not always equal to 8.3###, it depends on the units you wish to use. Here, on the right is a table of the gas constant in other units:
http://en.wikipedia.org/wiki/Gas_constant
« Last Edit: June 22, 2007, 12:20:00 AM by enahs »

Offline Ryan

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Re: question about this aqueous solution problem
« Reply #4 on: June 22, 2007, 11:49:25 AM »
The total pressure would be quadrupled, the partial pressure increase due to the addition of Ar would be three times the initial pressure, but the final pressure includes the partial pressure contribution of the Ne as well.

Total pressure = 2 ATM

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