What is the concentration of H2 for the equation H2+I2=2HI if the equilibrium constant is 64 and the initial concentrations are both 1.9
let the amount of $\ce{H2}$ reacted = $\pu{x~mol}$ 
Use the molar ratio to detemine the amount of $\ce{I2}$ reacted and the amount of $\ce{HI}$ formed:
$$\begin{align}
&amp;\text{The amount of }\ce{H2}~\text{reacts to reach equilibrium }&amp;=\pu{x~mol}\\
&amp;\text{The amount of }\ce{I2}~\text{reacts to reach equilibrium}&amp;=\pu{x~mol}\\
&amp;\text{The amount of }\ce{HI}~\text{formed at equilibrium }&amp;=\pu{2x~mol}\\
\end{align}$$
Use the following table to determine the amount and the concentration 
of each species at equilibrium : 
$$\begin{align}
\ce{&amp;H2 &amp;&amp;+ &amp;&amp;O2 &amp;&lt;=&gt; &amp;&amp;2HI\\
I~~~~  &amp;1.9   &amp;&amp;&amp;&amp;1.9      &amp;&amp;&amp;0 \\
C ~~~  &amp;-x    &amp;&amp;&amp;&amp;-x       &amp;&amp;&amp;2x \\
E ~~~  &amp;(1.9-x) &amp;&amp;&amp;&amp;(1.9-x) &amp;&amp;&amp;2x} 
\end{align}$$
Th concentrations at equilibrium :
$$[\ce{H2}]_\mathrm{e}=(1.9-x)  , [\ce{I2}]_\mathrm{e}=(1.9-x) ,  [\ce{HI}]_\mathrm{e}= \pu{2x~M} $$
Sustitute the concentrations in the following formula:
$$\begin{align}
K_\mathrm{C} = \frac{ [\ce{HI}]_\mathrm{e}   }  {[\ce{H2}]_\mathrm{e}[\ce{I2}]_\mathrm{e} }\\
64= \frac{(2x)^2}{(1.9-x)(1.9-x)}
\end{align}$$  
Take the square root of both sides :
$$\sqrt{64}=\sqrt{\frac{(2x)^2}{(1.9-x)^2}}$$


$$8=\frac{2x}{(1.9-x)}$$
- Solve for $x : x=1.52$


Calculate the concentrations at equilibrium :
$$\begin{align}
[\ce{H2}]_\mathrm{e}&amp;=(1.9-x)&amp;=(1.9-1.52)&amp;=\pu{0.38~M}  \\
[\ce{I2}]_\mathrm{e}&amp;=(1.9-x)&amp;= (1.9-1.52)&amp;=\pu{0.38~M} \\ 
 [\ce{HI}]_\mathrm{e}&amp;= \pu{2\times{x}}&amp;= 2\times{1.52}&amp;=\pu{3.04~M}
\end{align} $$