Since we are already given the balanced equation:
[tex] Br_{2}+2NaI [/tex] → [tex] NaBr+I_{2} [/tex]
We can derive the molar ratios as: 1:2:1:1
That being said, we are given 0.172 moles of bromine ([tex] Br_{2} [/tex]), so it has a ratio of 1:1 with sodium bromide ([tex] NaBr [/tex]).
So we can take from that ratio, that when 0.172 moles of bromine are used, we are, in turn, going to get 0.172 moles of sodium bromide produced.