Mg + 2H+ --> Mg2+ + H2
A student performed an experiment to determine the amount of hydrogen gas released in a reaction. The student produced the hydrogen gas by reacting HCL and a strip of Mg metal according to the equation above. All of the Mg metal was consumed and the hydrogen gas was collected by displacement of water in an inverted bottle. The student’s data contain the following information:
Mass of Mg => 0.024g
Volume of gas collected over water => 25.2mL
Water Temp => 22C
Room Temp => 22C
Atmospheric Pressure => 749.8 mm Hg
Vapour pressure of water at 22C => 19.8 mm Hg
The volume of the dry hydrogen gas at 1 atm and room temp would be :
A [(25.2)(749.8+19.8)]/760
B [(25.2)(760-19.8)]/749.8
C [(25.2)(749.8-19.8)]/760
D (749.8-16)/[(760)(25.2)]
E (760-16)/[(749.8)(25.2)]
I have no idea how to proceed with this. Any help would be appreciated. The answer is C
I’m a bit rusty. First, start with what you know. All Mg consumed=mg is the limiting rxt.
Now, for the subject tests they tend not to “make” you do the actual math. However the pain in the tush is setting it up ow they want.
**=step I would take as a standard if unsure but may not be necessary
**Convert Mg to moles. (I won’t do this here yet because I’m not sure it’s relevant)
**When you get atm subtract vapor pressure of water (hence 760-19.8
I think this may be as simple as P1V1=P2V2
So let initial volume be V1, P1 being 760.
Why 760? You’re in mmHg at 1 atm .: 1atm=760mmHg
Let V2=collected gas=25.2ml, let P2= subtracted amount of collected gas
So! V1(760)=(25.2)(749.8-19.8)
Then divide and intial volume = (25.2)(749.8-19.8)/(760)
side tip, if you ever get a Celsius number eliminate answers with it instead of c+273 because we work in K not C (just a common trick I’ve noticed)
ALSO indirect variation (and direct/proportions) are your BEST friends. This and PV=nRT are your best friends.
If this doesn’t make sense, shoot me a message and I’ll try to do a better job. It’s late (ironically, I’m doing bio) but I’ll do my best ^^
Thank you so much that was really helpful 
If you collected hydrogen gas by the displacement of water and under the conditions shown: https://s13.postimg.org/st6ge7tqf/ssa.jpg (link to image) , which of the following would give you the
pressure of the hydrogen in the bottle?
(A) 730. mm − 40.8 mm
(B) 730. mm − 30.0 mm
(C) 730. mm − 30.0 mm/13.6 + 40.8 mm
(D) 730. mm − 30.0 mm/13.6 − 40.8 mm
(E) 730. mm − 40.8 mm/13.6 − 30.0 mm
This similar to what i asked before but I’m not quite sure what 13.6 is supposed to be here. Is it the density of mercury?
The answer is E. Appreciate your help.
I’m sorry I can’t help with this one. Are you sure there isn’t more info? Either way don’t let it get you down. Just do your best tomorrow and save toughies like this for the end
