Are you sure this is the equation?
(CO) = 0.25/15 L = about 0.0167M
(H2) = 0.90/15 L = about 0.06 M
........CO + H2 ==> H2O + C
I....0.0167.0.060....0.....0
C.......-x...-x.......x....
E.....0.0167-x.0.060-x..x
(H2O) = 0.190/15 L = about 0.0127 = x
Then 0.0167-x = ?
and 0.060-x = ?
Substitute into Kc expression and solve for Kc.
0.25mol of CO are mixed with 0.90mol of H2 and heated to 950 degrees celcius in a 15L flask at equilibrium 0.190mol of H2O were present, what is the concentration of the other reactants and products at equilibrium and what is Kc
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