........CO(g) + Cl2(g)⇌ COCl2(g)
E.....1.9E-6...9.3E-7.....x
Substitute the E line into the Keq expression and solve for x = (COCL2) in mols/L.
Then mols/L x 5.89 L = mols COCl2.
Grams COCl2 = mols COCl2 x molar mass COCl2.
Consider the reaction:
CO(g)+Cl2(g)⇌COCl2(g)
Keq= 2.9×1010 at 25 ∘C
A 5.89 −L flask containing an equilibrium reaction mixture has [CO]= 1.9×10−6 M and [Cl2]= 9.3×10−7 M .
How much COCl2 in grams is in the equilibrium mixture?
2 answers
Keq expression for the following: Co + Cl2↔CoCl2