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 Post subject: Molar Volume Problem?
PostPosted: Tue Apr 22, 2014 8:49 am 
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A reaction of Mg and HCl generates 20.0 mL of gas, which was collected by water displacement in a 22 degrees celcius bath. The barometric pressure was 740 mmHg.

Use the combined gas law to calculate the volume of hydrogen at STP.

Attempt at solution:
I calculated that the pressure of the hydrogen gas is 720 mmHg, the temperature is 293 K, but how do I figure out the volume of the hydrogen gas? Do I just use 20.0 mL? However, isn"t some of that 20.0 mL water vapor? Once I can figure this out I know it is simply plugging in the values into P1V1/T1 = P2V2/T2.

Thanks.


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 Post subject: Molar Volume Problem?
PostPosted: Sat Aug 05, 2017 6:43 pm 
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here the idea. That gas you collected has some H2 in it and some H2O. Because youre collecting it over water.

and how we account for that i thi. We assume the volume of H2 = volume H2O = 20.0 mL. The accounting difference is in the PRESSURE.

Ptotal = PH2 + PH2O

and PH2O i the vapor pressure of water at the given temperature. You have to look that up.

*************
from here..
http://en.wikipedia.org/wiki/Water_(data_paje)
in the table at the bottom of the page
the vapor pressure of water at 22°C = 26.4442 hPa

and yes.. those are weird units so let convert hPa to mmHg
26.4442 hPa x (100Pa / 1hPa) x (760mmHg / 101325 Pa) = 19.835mmHg

then..
PH2 = 740mmHg - 19.835mmHg = 720mmHg

then we can use...
P1V1 / T1 = P2V2 / T2
to find V2

V2 = V1 x (P1 / P2) x (T2 / T1)
V2 = 20.0mL x (720mmHg / 760mmHg) x (273.15K / 295.15K)
V2 = 17.5mL

**********
questions about any of this?

*********
update.. NO.. it doesnt mean I produced the same amount of H2O or twice as much...

think of the ga in the 20.0mL. it contain some H2 molecules bouncing around. They collide with the container. Their impact cause a pressure. Now think of the water molecule. Also bouncing around the same container. They collide with the container and also produce a pressure. Both H2 and H2O are contained in the 20.0 mL volume. So both gases have V = 20.0mL

Now imajine adding a third gas. 1 molecule of CO2. that would have a volume = 20.0mL. But not much pressure because you only have 1 molecule bouncing around. Now imajine you have a lot of CO2. Still contained in the 20.0mL. But now all those molecules are bouncing around adding pressure.

the total pressure = pressure from H2 collision + pressure from H2O collision + pressure from CO2 collision. Right?

and thi may help to.. pressure = force / area. force = mas x acceleration. a molecule of H2 or H2O or CO2 has mas. and when it hit the wall, it velocity i reversed. So it must decelerate and then accelerate. meaning it hit the wall with a force. sum up the force, divide by area and you get pressure.



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