Yes; however much is consumed of the one, the same must be consumed of the other. Also, the same amount of product must be formed (assuming they all react on a 1-to-1 ratio; that is, if the coefficients are the same. If not, then the amount used by a substance with a coefficient of 1 is multiplied by the coefficient of each substance to find the amount used/produced of that substance. See second paragraph). This is true no matter if the reaction proceeds to equilibrium or if it stops at a midway equilibrium (everything does, of course, but some go so close to completion as to make no difference). In this case, of course, because we are dealing with an equilibrium, and the reaction does not go to completion; since .8 mol product were formed, .8 mol of Cl2 was used, and .8 mol CO as well.
And if the formula were somehow 2CO+Cl2, then two moles of CO would be needed for every mole of Cl2. So, 3 moles would be used; or rather, 1.6, because of equilibrium.
Because of this, if in the first case there was 2 mol CO and 3 mol Cl2, then at the end there would be 2 mol COCl2 and 0 mol CO, and there would remain 1 mol Cl2. That 1 Cl2 has nothing to react with, and so it hangs around (according to the formula, at least). Thus, CO “limits” the reaction; it is called the limiting reagent. In the second case I gave, even if there were 3 mol CO and 2 mol Cl2, CO is still the limiting reagent, because it takes 2 moles of CO to react with 1 mole of Cl2, so if 2 mol Cl2 can form 2 mol product, 3 mol CO can only form 1.5. So to find the limiting reagent, just calculate the amount of product that each reactant can possibly form, given their amounts, and take the lowest one.