Percent composition by volume of the reaction H2 + Br2  2HBr For the reaction $\ce{H2(g) + Br2(g) &lt;=&gt; 2HBr(g)}$ at a certain temperature, $K=1$. Determine the composition (in per cent by volume) of the equilibrium reaction mixture if the initial mixture consisted of $3$ moles of $\ce{H2}$ and $2$ moles of $\ce{Br2}$.


The answer given is $49.6\%$ of $\ce{H2}$, $29.6\%$ $\ce{Br2}$, $20.8\%$ $\ce{HBr}$.

I made an ICE table with following data:

$$
\begin{array}{c|ccc}
   *        &amp; \mathbf I   &amp; \mathbf C   &amp; \mathbf E \\
\hline
 \ce{H2}    &amp; 3   &amp; -x  &amp; 3-x \\
 \ce{Br2}   &amp; 2   &amp; -x  &amp; 2-x \\
 \ce{HBr}   &amp; 0   &amp; 2x  &amp; 2x \\
\hline
\end{array}
$$

Using this data I made the equation:

$$
1=\frac{4x^2}{(3-x)(2-x)}
$$

My work in solving it was:

$$
(3-x)(2-x)=4x^2 \\
3x^2+5x-6=0 \\
x=\frac{-5+9.85}{6}\text{ or } x=\frac{-5-9.85}{6}
$$

I used the answer $0.808$ for $x$. I calculated the volume percents as:

$$
\%~\ce{H2} = {3-x\over 5} = 43.84\% \\
\%~\ce{Br2} = {2-x\over 5} = 23.84\% \\
\%~\ce{HBr} = {2x\over 5} = 32.33\%
$$

But, these don't match the answer given. Where have I gone wrong?
Your equation and your math appear correct to me: I also get $x=0.808$ from the quadratic equation, and calculate the same percentages.

As long as you are supposed to assume ideal gas behavior in the question, it appears to me that the answer key is incorrect.