How does nitric acid reaction with sodium hydroxide?

$H N {O}_{3} \left(a q\right) + N a O H \left(a q\right) \rightarrow N a N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$
Moles of nitric acid: $680 \times {10}^{- 3} \cancel{L} \times 0.640 \cdot m o l \cancel{{L}^{- 1}}$ $=$ ??? $m o l$. I included the units in the calculation, because if a calculation makes sense dimensionally (and here it did, I required an answer in $m o l$), it is a good indication that I have set up the problem correctly.