For the reaction cocl2(g)⇌co(g)+cl2(g), k= 2.19×10−10 at 373 k
what can be said about this reaction at this temperature?

(a) the equilibrium lies far to the right.
(b) the reaction will proceed very slowly.
(c) the reaction contains significant amounts of products and reactants at equilibrium.
(d) the equilibrium lies far to the left.

D) The equilibrium lies far to the left

Explanation:

According to the law of mass action, the equilibrium constant K for the reaction at 373K can be calculated as follows:

K = = 2.19×10^{-10}

([X] means = concentration of X)

This means that in the equilibrium the concentration of the reactant (that is in the denominator) will be much higher (around 10^{10} fold) than the concentrations of the products (that are in the numerator), and this means that the equilibrium lies far to the left (to the reactants side) as very small amount of product is being formed.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

what does the table look like