Answer
$$2HClO_4(aq) + Li_2CO_3(aq) \longrightarrow CO_2(g) + H_2O(l) + LiClO_4(aq)$$
Work Step by Step
1. This reaction has a carbonate and an acid as its reactants. Thus, it will produce $CO_2(g)$, $H_2O(l)$ and a salt made of the other ions:
$$HClO_4(aq) + Li_2CO_3(aq) \longrightarrow CO_2(g) + H_2O(l) + Salt$$
The remaining ions are: $Li^{+}$ and $Cl{O_4}^-$, therefore, the salt is $LiClO_4$: $$HClO_4(aq) + Li_2CO_3(aq) \longrightarrow CO_2(g) + H_2O(l) + LiClO_4(aq)$$
2. Now, balance the equation.
To balance the number of lithium atoms, put a $2$ before $LiClO_4$
$$HClO_4(aq) + Li_2CO_3(aq) \longrightarrow CO_2(g) + H_2O(l) + 2LiClO_4(aq)$$
To balance the number of chlorine atoms, to the same for $HClO_4$:
$$2HClO_4(aq) + Li_2CO_3(aq) \longrightarrow CO_2(g) + H_2O(l) + 2LiClO_4(aq)$$
