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?
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.
..........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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.