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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: dado on January 25, 2009, 02:10:03 PM

Title: Chemical exercises
Post by: dado on January 25, 2009, 02:10:03 PM
Title: Re: Chemical exercises
Post by: Dan on January 25, 2009, 03:07:03 PM
Title: Re: Chemical exercises
Post by: dado on January 26, 2009, 05:41:43 AM
Title: Re: Chemical exercises
Post by: sjb on January 26, 2009, 07:30:54 AM
Title: Re: Chemical exercises
Post by: dado on January 26, 2009, 04:09:42 PM
can you tell me how did you solve this and what are your results
Title: Re: Chemical exercises
Post by: sjb on January 27, 2009, 06:45:25 AM
Having reread what you've put, this looks OK. It seems a bit odd having the equation for water formation as 2H + 1/2(g) + O2(g)  :rarrow: H2O(g), but if that's what it says, fair play.

I got a little confused between O (the letter between N and P) and 0 (the number zero)

I'd make sure you have the right enthalpy of formation for water though, I don't have my data books to hand, but I seem to recall there was data for liquid water and for gaseous water as well.

However, for the first question, I don't see any indication that you've corrected the volume of hydrogen generated to standard conditions (though it is possible I suppose that the two altered conditions cancel). What is the volume of one mole of gas under your conditions? (pV = nRT)

S
Title: Re: Chemical exercises
Post by: sjb on January 27, 2009, 08:45:23 AM
Quote from: sjb on January 27, 2009, 06:45:25 AM
It seems a bit odd having the equation for water formation as 2H + 1/2(g) + O2(g)  :rarrow: H2O(g)

That should of course read "2H(g) + 1/2O2(g)  :rarrow: H2O(g)", sorry

S
Title: Re: Chemical exercises
Post by: Dan on February 01, 2009, 02:00:34 PM
Quote from: dado on January 26, 2009, 04:09:42 PM
can you tell me how did you solve this and what are your results

Ok, it seems we calculated the number of moles of hydrogen differently, here's how I did it:

p = 0.98 bar
V = 52.8 dm3
T = 293 K
R = 8.31 x 10-2 dm3 bar K-1 mol-1

pV = nRT

n = pV/RT = 2.13 mol

Mass of Calcium Hydride needed = 0.5 x 2.13 mol x 42.1 g mol-1 = 44.8 g

Title: Re: Chemical exercises
Post by: Borek on February 01, 2009, 02:42:35 PM
;)