$9$

$10$

The equilibrium constant $K_c$ for the reaction

$H_2(g)+B{r}_2(g)2HBr(g)$

is $2\mathrm{.}180{10}^6$ at $730\mathrm{.}0C.$ Starting with $1\mathrm{.}20$ moles of $HBr$ in a $17\mathrm{.}8L$ reaction vessel, calculate the concentrations of $H_2,B{r}_2,$ and $HBr$ at equilibrium.

Part: $\frac{0}{2}$

Part $1$ of $2$

Round each of your answers to $3$ significant digits.

$[H_2]=,M$

$[B{r}_2]=,M$