Answer
$$2HBr(aq) + Na_2S(s) \longrightarrow H_2S(g) + 2NaBr(aq)$$
Work Step by Step
1. This reaction has a sulfide and an acid as its reactants. Thus, it will produce $H_2S(g)$ and a salt made of the other ions:
$$HBr(aq) + Na_2S(s) \longrightarrow H_2S(g) + Salt$$
The remaining ions are: $Na^{2+}$ and $Br^-$, therefore, the salt is: $NaBr(aq)$
$$HBr(aq) + Na_2S(s) \longrightarrow H_2S(g) + NaBr(aq)$$
2. Now, balance the equation.
To balance the number of sodium atoms, put a $2$ before $NaBr$
$$HBr(aq) + Na_2S(s) \longrightarrow H_2S(g) + 2NaBr(aq)$$
The number of $H$s and $Br$s on the products side is equal to 2, thus, we just need to put a $2$ before $HBr(aq)$:
$$2HBr(aq) + Na_2S(s) \longrightarrow H_2S(g) + 2NaBr(aq)$$
