Decrease of volume by half in equilibrium - with specific numbers- how to solve?


> 
> $$\ce{CO (g) + Cl2(g)<=>COCl2}(g)$$
>  We have
>  2 mols of $\ce{CO}$, 1 Mol of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
>  contained in 1 L.
>  The system is currently in equilibrium. 
> 
> 
> We then decrease the volume by half to 0.5 L
> 
> 
> What will be the new concentration in equilibrium of CO?
> 
> 
> 


I tried calculating at first the constant of concentration; 


$Kc=\frac{[\ce{CoCl2}]}{[\ce{CO}]\times[\ce{Cl2}]}$


I found it to be $4.8825$


then I calculated the concentration with the new Volume:


$[\ce{CO}]= \pu{2mol}/\pu{0.5L}= \pu{4 M}$


$[\ce{Cl2}]= \pu{1mol}/\pu{0.5L}= \pu{2 M}$


$[\ce{CoCl2}]= \pu{9.765mol}/\pu{0.5L}= \pu{19.53 M}$


obviously this wasn't in equilibrium so I went to do this:


$\ce{CO + Cl2<=>COCl2}$


$4M-x$, $2M-x$, $19.53+x$


(as the number of mols in the products is smaller and the system wants to fix the increase in pressure) 


When I calculated all of this in the Kc formula it came to me that the Concentration of CO increased, and is now larger than 2 M , 
and in the textbook answers it is stated that it is supposed to be less than 2M, how is it possible? 


I would be very glad for help with calculations and technique of solving as well!




I solved the quadratic equation by [Wolfram Alpha](https://www.wolframalpha.com/input/?i=4.8825*X%5E2-30.295*X%2B19.53%3D0)


[![enter image description here](https://i.stack.imgur.com/ycfdQ.png)](https://i.stack.imgur.com/ycfdQ.png)


The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}<\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}<\pu{2M}$, if we have
**1 mol** of $\ce{CO}$, **2 Mols** of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
contained in 1 L.


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