Asked by Ade

A mixture of 2.5 moles of H2O and 100 g of C are placed in a 50.0 L container and
allowed to come to equilibrium in the following reaction:
C(s) + H2O(g)  CO(g) + H2(g)
The equilibrium concentration of hydrogen gas is found to be 0.040 M. What is the
equilibrium concentration of the water vapour?

Answers

Answered by DrBob222
(H2O) = 2.5 mol/50 L = 0.05 M
..........C + H2O ==> CO + H2
I.......solid.0.05.....0....0
C.......solid -x.......x....x
E......solid.0.05-x....x....x

Kc = (CO)(H2)/(H2O)
The problem give you H2, CO is the same H2, you plug in Kc and solve for (H2O)
You didn't give Kc in the problem and there isn't enough information to calculate it.
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