Which coefficient before sodium bromide (NaBr) balances this chemical equation?

2NaF + Br2 → __NaBr + F2

We see that in the left-hand side of the equation, the side of the reactants, that we have 2 moles of Na and bromine is in it's diatomic form.

Therefore, we have 2 moles of each of these elements. When we combine the sodium bromide molecule in the products, we are going to want to keep these molar amounts the same. So we are going to need to put a 2 in front of the sodium bromide in order to correctly balance this equation.

So the coefficient for sodium bromide (NaBr) in this equation is 2.