I need help on how to set this up?

Question

N2O gas was generated from the thermal decomposition of ammonium nitrate and collected over water. The wet gas occupied 126mL at 21C when the atmospheric pressure was 758 torr. What volume would the same amount of dry N2O have occupied if collected at 758 torr and 21C? the vapor pressure of water is 18.65 torr and 21C.

this is what i setted up so far
P1= .99atm P2= .025 atm
T1= 294 K T2=294K
V1= .126 L V2=?

I know we have to set this up in this eqn
P1V1 / P2V2 = N1R1T1 / N2R2T2
since R is constant that is cancelled...but what about N?

Answer 1

u did a good job. u realized that this was a

P1V1/n1T1 = P2V2/n2T2. R cancelled like u said. n cancels since it says in the problem "what would the same amount of N2O occupy." this means that the moles stayed constant. u could also cancel T since the temperature didnt change. now this is a boyles law problen. now, u need not convert torr to atm nor L to mL as long as ur consistant.

P1V1 = P2V2
(758)(126) = 18.65 V2

5121 mL or 5.121L = V2

Answer 2

21C=294K
758torr - 18.65torr = 739torr

P1V1 = P2V2

V2 = P1V1/P2

V2 = (739)(126)/(758) = 123 mL to three sig figs.