Asked by qwerty
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?
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?
Answers
Answered by
DrBob222
Have you checked these numbers and the post? Have you checked the narrative?
Answered by
Alexis
Someone please answer this. Please! :)
Answered by
Mike
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
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