Calculating [HI] given Kc, [H2], and [I2] for the reaction 2HI  H2+I2 In the reaction in equilibrium
  
  $$\ce{2HI &lt;=&gt; H2 + I2}$$
  
  The value $K_c$ is $0.0198$ at $\pu{721 K}$. Calculate $\ce{[HI]}$ at $\pu{721 K}$ if $\ce{[H2]} = \pu{0.0120 M}$ and $\ce{[I2]} = \pu{0.0150 M}$.


We use an ICE box. Since $K &lt; 1$, the equilibrium is to the left. We are spending products to create reagents, so...

$$\begin{bmatrix}&amp;\ce{[HI]} &amp; \ce{[H2]} &amp; \ce{[I2]}\\
\text{Initial} &amp; 0 &amp; 0.0120 &amp; 0.0150\\
\text{Change} &amp; +2x &amp; -x &amp; -x\\
\text{Equilibrium} &amp; 2x &amp; 0.0120 - x &amp; 0.0150 - x
\end{bmatrix}$$

The equilibrium is given by

$$0.0198 = \frac{(0.0120-x)(0.0150-x)}{(2x)^2}$$

Solving yields:

$$x = 0.01907$$

Therefore

$$\ce{[HI]} = \pu{0.0381 M}$$

But the options are


  $\pu{0.00211M}$
  
  $\pu{0.133M}$
  
  $\pu{0.0334M}$
  
  $\pu{0.006567M}$
  
  $\pu{0.0953M}$


While it is pretty close to $\pu{0.0334M}$, I can't see how could I have made my calculation any more accurate (or if they were the ones who were inaccurate).

Was my procedure correct?
You're close. Using the expression for $K_c$, you just need to solve for $[\ce{HI}]$:

$$K_c = {[\ce{H2}][\ce{I2}]\over[\ce{HI}]^{2}}$$

$$0.0198 = {(\pu{0.012 M})(\pu{0.015 M})\over[\ce{HI}]^{2}}$$

$$[\ce{HI}] = \sqrt{{(\pu{0.012 M})(\pu{0.015 M})\over 0.0198}} = \pu{0.0953 M}$$