Posted by **plzprovidetheanswer** on Friday, October 4, 2013 at 9:55pm.

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)

- Chemistry -
**DrBob222**, Friday, October 4, 2013 at 10:20pm
This one may be a little different.

- Chemistry -
**plzprovidetheanswer**, Saturday, October 5, 2013 at 12:04am
i have no clue on how to set this up....

- Chemistry -
**DrBob222**, Saturday, October 5, 2013 at 12:53am
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....8.213-x..16.43-2x..x...4x

And the problem tells you (CS2) = 0.04107 mols/L so that x 130.0 = mols and you go from there.

- Chemistry -
**plzprovidetheanswer**, Saturday, October 5, 2013 at 1:01am
.04107 x 130 = 5.3391

4x= 5.3391

x = 1.3348

therefore H2S(g) = 1.3348/130

= .01027?

- Chemistry -
**DrBob222**, Saturday, October 5, 2013 at 1:10am
.04107 x 130 = 5.3391 **OK here**

4x= 5.3391 ** Why multiply by 4. It's H2S you want, not H2. So H2S mols = 16.43-2x = 16.43 -(2*5.339) = ? and then (H2S) = mols/L.**

x = 1.3348

therefore H2S(g) = 1.3348/130

= .01027?

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