Balance the following equation.
$K B r + C l_{2} ⟶ K C l + B r_{2}$

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Solution

Atoms of $C l$ in $R H S = 1$ & $L H S = 2$
Atoms of $B r$ in $R H S = 2$ & $L H S = 1$
Hence, we need to balance the $C l$ & $B r$ atoms. It can be balanced by taking $2$ moles of $K B r$ , which will give $2$ moles of $K C l$ & $1$ mole of $B r_{2}$ .
$2 K B r + C l_{2} ⟶ 2 K C l + B r_{2}$

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