In the reaction in equilibrium
$$\ce{2HI} \leftrightarrow \ce{H2 + I2}$$
The value $K_c$ is $0.0198$ at $721$ K. Calculate $\ce{[HI]}$ at $721$ K if $\ce{H2} = 0.0120$ M and $\ce{I2} = 0.0150$ M.
We use an ICE box. Since $K < 1$, the equilibrium is to the left. We are spending products to create reagents, so...
$$\begin{bmatrix}&\ce{[HI]} & \ce{[H2]} & \ce{[I2]}\\ \text{Initial} & 0 & 0.0120 & 0.0150\\ \text{Change} & +2x & -x & -x\\ \text{Equilibrium} & 2x & 0.0120 - x & 0.0150 - x \end{bmatrix}$$
The equilibrium is given by
$$0.0198 = \frac{(0.0120-x)(0.0150-x)}{(2x)^2}$$
Solving yields:
$$x = 0.01907$$
Therefore
$$\ce{[HI]} = 0.0381$$
But the options are
$0.00211$
$0.133$
$0.0334$
$0.006567$
$0.0953$
While it is pretty close to $0.0334$, 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?