Question

An equilibrium mixture contains 0.250 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container.

CO(g) + H2O(g) <-> CO2(g)+ H2(g)

How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answer 1

Have you checked these numbers and the post? Have you checked the narrative?

Answer 2

Someone please answer this. Please! :)

Answer 3

Kc= (.250)(.250)/(.200)(.200) = 1.56
CO + H20 --> CO2 + H2
.200 .200 .250 .250
.300 .300 .150+x .150

1.56 = (.150+x)(.150)/(.300)(.300)
Solve for x
X=0.788 moles