Determine the equilibrium constant for this reaction The question is:


  Ions $\ce{B}$ and $\ce{C}$ react to form complex $\ce{BC}$. If $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{B}$ is combined with $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{C}$, and our initial concentration of BC is $0$, yet the end in equilibrium for complex $\ce{BC}$ is $\pu{0.00450 mol}$ determine the equilibrium constant for this reaction.


Firstly, I converted the total volume $\pu{15.0 mL + 15.0 mL = 30 mL}$ to L and I got $\pu{0.03L}$

Secondly, I used the given $\pu{0.00450 mol}$ and divided it by $\pu{0.03 L}$ and I got $\pu{0.15 M}$ for BC 

Thirdly, I did the ICE chart



After that, I used the quadratic formula to solve for $x$ and calculate the concentration for each substance, as well as getting the $K_\mathrm{c}$ at equilibrium



I got $1.53$ for $K_\mathrm{c}$ but I am not sure if the whole process is correctly done. Please point out my mistakes
You made a slight error in calculations for B and C. While those concentrations are correct, that is 1 M at 15 mL. When doing RICE tables and equilibrium calculations, we can use concentrations, however I prefer to convert concentration into moles.

We are trying to find $K_c$, and we know that equal moles of the reactant react to give us 0.0045 mol of BC.

$$\ce{B + C &lt;=&gt; BC}$$

\begin{array} {|c|c|c|c|}
\hline
\text{Initial conc.} &amp; \text{0.015 mol} &amp; \text{0.015 mol} &amp; \text{0 mol}\\
\hline
\text{Change conc.} &amp; -x &amp; -x &amp; +x\\
\hline
\text{End conc.} &amp; \text{0.015 mol} - x &amp; \text{0.015 mol} - x &amp;  0 + x \\
\hline
\end{array}

We know that our change in equilibrium is 0.00450 mol, as that is the end amount of product that we have in our equilibrium calculations.

Accounting for the concentrations, we divide our number of moles by the total volume to get mol/L. 0.5 mol/L, 0.5 mol/L, .15 mol/L. 

Assuming that our reaction is at equilibrium:

$$K_c = \frac{[\text{0} + \text{.15 mol/L}]}{[\text{0.5 mol/L} - \text{.15 mol/L}][\text{0.5 mol/L} - \text{.15 mol/L}]} $$

$$K_c = \frac{[\text{.15 mol/L}]}{\text{[0.35 mol/L]}^2} $$

$$ K_c = 1.22 $$

Thanks @Nicolau for pointing out the error in the question and reviewing my work.