At certain $T$, $K_C$ for $H_2\left(g\right)+I_2\left(g\right)\rightleftarrows HI\left(g\right)$ is 16. If the molar concentrations of $HI\left(g\right)$ and $I_2\left(g\right)$ respectively are 0.4 M and 0.1M, then $\left[H_2\right]$ at equilibrium in mol/lit.

The balanced chemical equation is

$H_2\left(g\right)+I_2\left(g\right)\rightleftarrows HI\left(g\right)$

We know that for the expression $K_C$ is as follows

$K_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}$

It is given that $K_C$ = 16 $I_2\left(g\right)$ = 0.1M 

Putting these in above equation we get,

$\Rightarrow 16=\frac{{\left(0.4\right)}^2}{\left[H_2\right]\left(0.1\right)}$

$\Rightarrow \left[H_2\right]=\frac{0.16}{16(0.(A)}$

$\left[H_2\right]=0.1$

Hence, the correct option is (A) 0.1.