Starting with o.3700 mol CO (g) and 5.400*10^-2 COCl2 (g), in a 3 liter flask at 668 K, how many moles of Cl2 will be present?
CO + Cl2 =++COCL2
KC = 1.2*10^3 at 668 K
How many moles of Cl2 will be present WHEN? Today? tomorrow? next week? perhaps you mean at equilibrium?
CO + Cl2==> COCl2
Set up an ICE chart.
initial:
CO = 0.37/3 = 0.123
Cl2 = 0.054/3 = 0.018
COCl2 = 0
change:
COCl2 = +x
CO = -x
Cl2 = -x
equilibrium:
CO = 0.123-x
Cl2 = 0.018-x
COCl2 = x
Kc = (COCl2)^2/(CO)(Cl2)
Substitute the equilibrium concns above into Ka expression and solve for x.