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### Topic: calculating grams given pressure, volume, and temperature  (Read 32743 times)

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#### laxplayer

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##### calculating grams given pressure, volume, and temperature
« on: November 05, 2007, 07:10:00 PM »
The problem states:
Calcium hydride, CaH2, reacts with water to form hydrogen gas.

CaH2(s) + 2H2O(l) Ca(OH)2(aq) + 2H2(g)

This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired. How many grams of CaH2 are needed to generate 18.0 L of H2 gas if the pressure of H2 is 770 torr at 17°C?

I have no idea how to even start this problem, so could someone point me in the right direction :]

#### Sev

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #1 on: November 05, 2007, 07:32:48 PM »
Just use PV=nRT to find mol of H2.  Work backwards for mol of CaH2.

#### laxplayer

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #2 on: November 05, 2007, 08:07:26 PM »
So there are .766 mol of H2 and .383 mol of CaH2?

For H2 I did (18)(1.0132)/(.0821)(290.15)

For CaH2 I divided .766 by two because of the mol ratio, I think that's what you're supposed to do..

#### constant thinker

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #3 on: November 05, 2007, 08:07:54 PM »
To expand on Sev's post...

PV=nRT is the ideal gas law. It's extremely useful in gas stoichiometry.
P= Pressure
V= Volume
n= moles (sometimes substitute with "m/M" m=mass of sample M=molecular weight)
R= a gas constant, and it depends on the units you are using. You will need to look this up, I'm sure wikipedia has them.
T= temperature in Kelvin (always Kelvin for gas equation and most mathematical equations because it does not have negatives, and you'll never hit 0 in a real world problem)

Your answer is right. I don't know where you pulled your numbers from to get .766 mols H2, but you came out with the right number of mols for CaH2
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#### laxplayer

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #4 on: November 05, 2007, 08:13:02 PM »
do you know if what I did is correct though?

#### constant thinker

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #5 on: November 05, 2007, 08:19:19 PM »
Here is how I did it...

PV=nRT solve for n so you can find the mols of H2
n= PV/RT
n=(770 torr * 18 L)/(62.364 * 290°K)
n=.766
.766 mol H2

Now multiple by the ratio for calcium hydride to hydrogen to get mols of calcium hydride.
.766 mol H2*(1 mol CaH2/2mol H2)
.383mol CaH2
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#### laxplayer

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #6 on: November 05, 2007, 08:23:59 PM »
oh, I did (1.0132 atm * 18 L)/(.0821 * 290 k)
Where did you get 62.364 from?

#### constant thinker

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #7 on: November 05, 2007, 08:28:14 PM »
That is the R value if you were to use torr. You did it right though, I see what you did. You converted the torr to atm.
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#### Borek

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##### Re: calculating grams given pressure, volume, and temperature
« Reply #8 on: November 06, 2007, 03:07:02 AM »
0.08205783 L*atm/(K*mol)
8,314472 L*kPa/(K*mol)
8.314472 J/(mol*K)
62,3637 L*mmHg/(K*mol)
83,14472 L*mbar/(K*mol)
1.987216 cal/(K*mol)
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