Chemical equilibrium

Tutorial 25

    

The equilibrium constant of the reaction $H_2(g)$ $+$ $I_2(g)$ $2HI(g)$ is $K_c$ $=$ $54$ at $426^oC$ If the molarities at equilibrium of $I_2$ and $HI$ were: $[I_2]$ $=$ $0.0013$ $\frac{mol}{L}$
$[HI]$ $=$ $0.016$ $\frac{mol}{L}$
what would then be $[H_2]$ at equilibrium

$54$ $=$ $\frac{(1.6\cdot 10^{-2})^2}{[H_2]\cdot 1.3\cdot 10^{-3}}$ $[H_2]$ $=$ $\frac{(1.6\cdot 10^{-2})^2}{54\cdot 1.3\cdot 10^{-3}}$ $=$ $3.6\cdot 10^{-3}\frac{mol}{L}$