Chemistry - Mole Concept Question with Solution | TestHub

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

Initial: $\frac{5x}{2}$$\frac{5x}{2}$ 5x

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

Initial: $\frac{x}{2}$$\frac{3x}{2}$ x

Total moles of ${\mathrm{Br}}_2$ reacted $=\frac{5x}{2}+\frac{x}{2}=3$

Total moles of ${\mathrm{Cl}}_2$ reacted $=\frac{5x}{2}+\frac{3x}{2}=4x$

Given that 3 moles of ${\mathrm{Br}}_2$ are reacted, we have:

$3x=3⟹x=1$

Therefore, the total moles of ${\mathrm{Cl}}_2$ reacted are:

$4x=4\times 1=4$