help! #1 and #3?

Question

#1. When heated, H2S(g) decomposes according to 2H2S(g) <=> 2H2(g) + S2(g) . A 3.40g of H2S(g) is introduced into an evacuated rigid 1.25L container. The sealed container is heated to 483K, and 3.72*10^-2 mol of S2(g) is present at equilibrium.
a) Write the expression for the equilibrium constant Kc for the decomposition rxn represented above.
b)Calc. the equilibrium concentration in mol L^-1 of H2(g) and H2S(g) at 483K.
c)Calc. Kc for the decomposition rxn at 483K.
d)Calc. partial pressure of S2(g) in the container at equilibrium at 483K.
e)For the rxn H2(g) + 1/2S2(g) <=> H2S(g) at 483K, calc. Kc.

#3. Answer the following about BeC2O4(s) and it's hydrate. a)Calc. the mass % of carbon in the hydrated form of the solid that has the formula BeC2O4 * 3H2O
b) When heated to 220'C, BeC2O4 * 3H2O(s) dehydrates completely. BeC2O4 * 3H2O -> BeC2O4(s) +3H2O(g)
If 3.21 of BeC2O4 * 3H2O is heated to 220'C, calc..
i) mass of BeC2O4(s) formed
ii) volume of H2O(g) released at 220'C and 735 mm Hg.

Answer 1

If x is the mole/liter of S2 (gas). We know it's value [ 0.0372 mole/1.25 Liter]. The equation tells us that there will be 2x moles/liter of H2, and that for each mole of H2 created, we lose one mole of H2S. We know the INITIAL H2S conc as [3.4 g/34 g/mole]/1.25L, and we know that we have lost 2x mole/liter at equilibrium.

a. The general expression for the equilibrium is
[H2]^2 [S2] / [H2S]^2 = Kc.
b. We did this above.
c. Stick the moles/liter of each species at equilibrium in the general equation.
d. For this, we need to sum all the moles in the 1.25 L container, and use the ideal gas law to compute the pressure at 483K. Then the partial pressure of S2 = (moles of S2)/(Total moles)xTotal pressure.
e. By comparison with the equation in a, this is a square root reduction (the terms squared become terms to the 1 power and the term to the 1 power is now a square root. So this Kc = sqrt(Kc) in (c) part of the problem.

3. a. Do your mol weight calculation for berillyium oxalate trihydrate. The mass % of carbon is 24 x100 divided by the mol wt of the trihydrate.
b. Since you know the mol weight of the trihydrate, you can find the moles in 3.21 g. This is the same moles as for the BeC2O4. So you multiply the moles by the anhydrous mol wt.
For each mole of the trihydrate, 3 moles of H2O are released. So you know the moles of H2O. Use the ideal gas law to find the volume for 220 deg C and 735 mm Hg.