[kemmy]

so i've been struggling with this one for hours...

C2H2 (acetylene gas) can be produced by reacting solid calcium carbide CaC2, with water. The products are acetylene and calcium hydroxide.

What volume of wet acetylene is collected at 25 degrees Celsius and 715 torr when 5.20 g of calcium carbide is reacted with an excess of water? (At 25 degrees Celsius the vapor pressure of water is 23.8 torr).

The back of the book says the answer is 2.26 L, but i dont know WHY. i used the ideal gas law equation, but perhaps im doing something wrong...

Thanks for any *delete me*

[Borek]

Show your work - especially how you dealt with the 'wet' part of the question.

[kemmy]

ok

so i used PV = nRT, which leads to V = nRT/P, since im trying to look for the volume.

for n (# of moles) i multiplied 5.20 g CaC2 to its molecular weight (64.1 g/mol) and got 0.0811 moles.

for T i got 298 K

for P i got 0.94 atm (converting 715 torr to atm).

the final answer i got was 2.11 liters.

im off by just a little bit, which makes the problem even more aggravating.

[Yggdrasil]

The wet acetylene gas consists of both acetylene molecules and water molecules.  The total pressure of the gas in 715 torr, but for this problem you need the partial pressure of acetylene in the gas.  Since you can assume that the gas you collect consists only of water and acetylene:

P_total = P_acetylene + P_water

where P_acetylene and P_water are the partial pressures of acetylene and water, respectively.

So, you should get the right answer when you use P_acetylene = 691.2 torr

[kemmy]

so i FINALLY figured it out. 

When i tried 691.2 torr, the final answer i got was 218 L, which is still not correct.

then i tried something new: i multiplied 23.8 torr by 2 because the problem said there was excess water, so i guessed that there would be 2 H2O molecules for every 1 C2H2.

well, 715 torr - (23.8 torr *2) = 667.4 torr, which is 0.878 atm.

and then i got the answer 2.26 L. 

[Borek]

You answer is correct only by coincidence. Partial pressures of both gases have nothing to do with the reaction stoichiometry - note that water is present in excess, it will be present in the gaseous phase even if will not take part in the reaction.

Use partial pressure definition to calculate total number of moles of both gases present, then convert this value to volume.

[kemmy]

how do i find the number of moles of water, if that's even possible? my guess is that i find the V for water and add it to the V for acetylene (adding up partial pressure equations). but without the moles i clearly cant find the V for water.

[Borek]

I told you - look out for partial pressure definition. It will let you find number of moles of H₂O.

[AWK]

Just neglect water using p=715-23.8 {[torr] - this is a pressure of dry acetylene.

[Borek]

Just neglect water using p=715-23.8 {[torr] - this is a pressure of dry acetylene.

Question asks about the volume of WET acetone.

[billnotgatez]

OK at 25C how much water vapor is in wet acetylene. Somehow I think this question is unfair. Also this Socratic method has me totally confused.

[kemmy]

i dont see how i can find the number of moles in water if i dont know anything about it except that there is an excess of it. and i did use the partial pressure equation only to find myself wondering what the volume of water is in this problem. if you can show me step by step how i can find the moles of water i'd really appreciate it.

this problem is in the zumdahl textbook SECOND edition (very old). yeah, the wording isnt very good in a lot of problems.

[Borek]

There is n_H2O and n_C2H2. Partial pressures are:

p_H2O = 23.8 = 715 * n_H2O/(n_H2O + n_C2H2)

p_C2H2 = 715 - 23.8 = 715 * n_C2H2/(n_H2O + n_C2H2)

Just solve for n_H2O.

[AWK]

Just neglect water using p=715-23.8 {[torr] - this is a pressure of dry acetylene.

This problem can be solved using different approaches. Borek calculates moles.
I suggest calculation a volume of pure acetylene then use proportionality of volumes and pressures

[billnotgatez]

CaC₂ + 2H₂O + excess H₂O = C₂H₂(g) + excess H₂O(v) + excess H₂O(l) + Ca(OH)₂(s)

I guess we assume that wet acetylene is the acetylene gas plus the water vapor.

I now see that the vapor pressure of water is given.

Writing out the equation makes me less confused.