sorry typo
PV=nRT
2n=PV/RT=(0.66atm)*(1L)/(0.0821 L-atm/mol-k)*(298K)=0.0269 mol
n=0.0135 mol
0.0135 mol * 51g/1mol NH4HS= 0.6879 g NH4HS .1 mol *51g/1 mol= 5.1 g
0.6879g/5.1g=0.135
0.135*100=13.5% dissociated
I accidentyl put an extra 0 in 0.06879 it should be 0.6879. The rest of the numbers...
Using the method I thought would work I got
PV=nRT
(0.66atm)(2L)=n(0.0821L-atm/mol-k)(298k)
n=0.5395
0.5395x51g/1mol= 2.752g of NH2HS to obtain equilibrium
does this seem right??
Homework Statement
I am having trouble with this question can anyone point me in the right direction?
Solid NH4HS dissociates according to the equation:
NH4HS (s) ---> NH3 (g) + H2S (g)
SOme solid NH4HS is placed into an initially evacuated vessel and when equilibrium is established there is...