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?
3 answers
Have you checked these numbers and the post? Have you checked the narrative?
Someone please answer this. Please! :)
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