${\mathrm{Br}}_2\left(\mathrm{l}\right)$reacts with ${\mathrm{Cl}}_2 \left(\mathrm{g}\right)$to form $\mathrm{BrCI}$and ${\mathrm{BrCl}}_3$ , simultaneously. How many moles of ${\mathrm{CI}}_2\left(\mathrm{g}\right)$ reacts completely with 3 moles of Br 2 (l) to give $\mathrm{BrCl}$and ${\mathrm{BrCl}}_3$ in 5 : 1 mole ratio?

$\frac{1}{2}{\mathrm{Br}}_2\left(l\right)+\frac{1}{2}{\mathrm{Cl}}_2\left(g\right)⟶\mathrm{BrCl}$

$\frac{5x}{2}$            $\frac{5}{2\mathrm{x}}$                       5x

$\frac{1}{2}{\mathrm{Br}}_2\left(l\right)+\frac{3}{2}{\mathrm{Cl}}_2\left(g\right)⟶{\mathrm{BrCl}}_3$

$\frac{\mathrm{x}}{2}$                  $\frac{3}{2}x$                   $\mathrm{x}$

$\frac{5x}{2}+\frac{x}{2}=3$    ${\mathrm{Cl}}_2=\frac{5}{2}x+\frac{3}{2}x$

3x =3                 4x

X =1         4x l = 4