for the reaction system H2 + X2 <->2HX Kc = 24.4 at 300 K a system made up fron these components which is at equilibrium contains 0.150 moles of H2 and 0.600 moles of HX in a 3.00 liter. calculate the number of moles of X2 present at equilibrium.
...........H2 + X2 ==> 2HX
equil...0.150.........0.600
Kc = 24.4 = (HX)^2/H2)(X2)
Can't you substitute for HX and H2 and solve for X2?
(HX) = 0.600 mol/3.00L = ?
(H2) = 0.150 mol/3.00L = ?
Solve for (X2) = ?M, then moles = M x L
They give answer 0.0984 mol and i get 3 something
ok iget 0.328 not 0.0984
0.0984 is right on the money. Post your work and I'll find your error.
HX) = 0.600 mol/3.00L = 0.2 M
(H2) = 0.150 mol/3.00L = 0.05M
(0.2)^2/0.05)(24.4)=0.327
ok i get 24.4/3 = 8.13 and the solve for x2
I don't know why you changed your post? Your 0.327 is ok if you didn't terminate that last number. You should have left it at 0.03279, then multiplied by 3.00 L.
Your last post makes no sense at all.
Kc = 24.4= (HX)^2/((H2)(X2)
(HX) = 0.600/3.00 = 0.200M
(H2 = 0.150/3.00 = 0.05M
Solve for (X2) = (HX)^2/(H2)(Kc)
(X2) = (0.200)^2/(0.05*24.4) = 0.03279M
moles X2 = M x L = 0.03279*3.00L = 0.09836 moles which rounds to 0.0984.