How to calculate the mass of phosgene in the equilibrium mixture of carbon dioxide and chlorine?


> 
> Consider the reaction:
> 
> 
> $$\ce{CO(g) + Cl2(g) <=> COCl2(g)}$$
> 
> 
> A closed $\pu{5.69 L}$ flask containing a reaction mixture at equilibrium has $[\ce{CO}] = \pu{1.8 \times 10^-6 M}$ and $[\ce{Cl2}] = \pu{6.3 \times 10^-7 M}$. Given that the equilibrium constant $K\_\mathrm{eq} = 2.9 \times 10^{10}$, calculate the mass of $\ce{COCl2}$ present in the flask.
> 
> 
> 


How can I solve the question above? I thought that equilibrium constants deal with concentrations and not masses, so I'm not sure how to use it to calculate a mass.




We have an equilibrium mixture for the reaction


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


We should recognize that


$$K\_\mathrm{eq} = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$


Note that the reactants and products all have stoichiometric coefficients of 1 so we can write our $K\_\mathrm{eq}$ equation out in this way.


Because we are given concentrations of reactants in the *equilibrium mixture*, we can go ahead and plug in the values that we know and solve for $[\ce{COCl2}]$.


$$\begin{align}
2.9 \times 10^{10} &= \frac{[\ce{COCl2}]}{(\pu{1.8 \times 10^-6 M})(\pu{6.3 \times 10^-7 M})} \\
[\ce{COCl2}] &= \pu{0.033 M}
\end{align}$$


We are asked how many *grams* of $[\ce{COCl2}]$ is present in the reaction mixture. Concentration and mass are related via the following equation:


$$[\ce{COCl2}] = \frac{n(\ce{COCl2})}{V} = \frac{m(\ce{COCl2})}{M(\ce{COCl2}) \cdot V}$$


or


$$m(\ce{COCl2}) = [\ce{COCl2}] \cdot M(\ce{COCl2}) \cdot V$$


where $n$ is the amount of $\ce{COCl2}$ (in g), $M$ is the molar mass (in g/mol), and $V$ is the volume (in L).


Let us plug in all of our known values and solve.


$$\begin{align}
m(\ce{COCl2}) &= (\pu{0.033 mol L-1})(\pu{98.906 g mol-1})(\pu{5.69 L}) \\
&= \pu{18.6 g}
\end{align}$$


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