[dado]

[Dan]

[dado]

[sjb]

[dado]

can you tell me how did you solve this and what are your results

[sjb]

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)  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

[sjb]

It seems a bit odd having the equation for water formation as 2H + ¹/₂(g) + O₂(g)  H₂O(g)

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

S

[Dan]

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

[Borek]