Question

How would the equilibrium concentration of `H_2O` be affected by removing `H_2` from the mixture?

Answer 1

All other things equal, there would be a decline in `[H_2 O]`.

Explanation:

The question is likely to be talking about the thermal decomposition of `H_2 O`, a nonspontaneous process that becomes reversible at ridiculously-high temperature. The chemical equation for this reaction- a reversible process- is:

`H_2 O (g) rightleftharpoons H_2 (g) + 1/2 color(white)(l) O_2 (g)`

Removing `H_2 (g)` reduces its concentration in the mixture (assuming that volume stays unchanged.)

The Le Chatlier's Principle predicts that the system would respond to this change in a way as if it's trying to minimize the change's impact on the equilibrium conditions. `H_2 O` would be consumed as the system try to produce `H_2` to (effortlessly) make up for the removal.

`n(H_2O)` declines whereas `V` stays the same (assumed.)
`c = n / V`, what would be the direction of the change in `c(H_2O)`?

Reference
"Water splitting." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 15 Jun. 2018. Web. 6 Jul. 2018, https://en.wikipedia.org/wiki/Water_splitting#Thermal_decomposition_of_water

"Thermochemical cycle." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 18 Mar. 2018. Web. 6 Jul. 2018, https://en.wikipedia.org/wiki/Thermochemical_cycle