Chemistry
posted by plzprovidetheanswer on .
When CH4(g) (0.06318 mol/L) and 16.43 mol of H2S(g) in a 130.0 L reaction vessel at 711.0 °C are allowed to come to equilibrium the mixture contains 0.04107 mol/L of CS2(g). What is the equilibrium concentration (mol/L) of H2S(g)?
CH4(g)+2H2S(g) = CS2(g)+4H2(g)

This one may be a little different.

i have no clue on how to set this up....

Note here that CH4 is given as mols/L (instead of mols for the other problems)
so if (CH4) = 0.06318 mols/L and the volume is 130.0L, then mols = 0.06318 x 130L = 8.213 mols.
......CH4 + 2H2S ==> CS2 + 4H2
I.....8.213..16.43....0.....0
C.....x.....2x......x.....4x
E....8.213x..16.432x..x...4x
And the problem tells you (CS2) = 0.04107 mols/L so that x 130.0 = mols and you go from there. 
.04107 x 130 = 5.3391
4x= 5.3391
x = 1.3348
therefore H2S(g) = 1.3348/130
= .01027? 
.04107 x 130 = 5.3391 OK here
4x= 5.3391 Why multiply by 4. It's H2S you want, not H2. So H2S mols = 16.432x = 16.43 (2*5.339) = ? and then (H2S) = mols/L.
x = 1.3348
therefore H2S(g) = 1.3348/130
= .01027?