Balance the following chemical equations:
$(a)$ $KClO_{3} \longrightarrow KCl + O_{2}$
$(b)$ $Na_{2}CO_{3} + HCl \longrightarrow NaCl + H_{2}O + CO_{2}$

(A) To balance a chemical equation,the number of atoms of each element must be equal on both the reactant and product sides.
$(a)$ For $KClO_{3} \longrightarrow KCl + O_{2}$:
There is $1$ $K$,$1$ $Cl$,and $3$ $O$ on the left,and $1$ $K$,$1$ $Cl$,and $2$ $O$ on the right. To balance oxygen,we multiply $KClO_{3}$ by $2$ and $O_{2}$ by $3$. Then,to balance $K$ and $Cl$,we multiply $KCl$ by $2$. The balanced equation is: $2KClO_{3} \longrightarrow 2KCl + 3O_{2}$.
$(b)$ For $Na_{2}CO_{3} + HCl \longrightarrow NaCl + H_{2}O + CO_{2}$:
There are $2$ $Na$ atoms on the left,so we multiply $NaCl$ by $2$. This gives $2$ $Cl$ atoms on the right,so we multiply $HCl$ by $2$. Checking the atoms: $2$ $Na$,$1$ $C$,$3$ $O$,$2$ $H$,and $2$ $Cl$ on both sides. The balanced equation is: $Na_{2}CO_{3} + 2HCl \longrightarrow 2NaCl + H_{2}O + CO_{2}$.