[laxplayer]

The problem states:
Acetylene gas, C2H2(g), can be prepared by the reaction of calcium carbide with water.

CaC₂(s) + 2 H₂O(l)----> Ca(OH)₂(s) + C₂H₂(g)
Calculate the volume of C₂H₂ that is collected over water at 23.0°C by reaction of 0.1580 g of CaC₂ if the total pressure of the gas is 766 torr?

So I found the vapor pressure of water to be 21.07 torr. I subtracted 21.07torr from 766torr to get 744.93torr as the pressure. This converts to .8902 atm.
I found the moles of CaC₂ by .1589 g CaC₂/64.1g to get .002465 mol
The temperature is 296.15 K.
I used the formula V=nRT/P, so I did:
(.002465 mol)*(.0821)*(296.15K)/.8902atm and got .06732622 L

Is this the right answer, or should I have found the moles of C₂H₂ instead, that's the only other way I could think of to do this problem.

[LQ43]

yes you should have calculated mol of C2H2 but you're lucky because in this case, mol of CaC2 = mol C2H2   WHY?

but otherwise, your calculation looks good

[laxplayer]

The coefficients are in a 1:1 ratio, so the moles are also