Answer
$8.32\ g$
Work Step by Step
Number of moles of nitrate:
$15.0\ g\div 85.00\ g/mol=0.176\ mol$
Number of moles of $NaNH_2$
$15.0\ g\div 39.01\ g/mol=0.384\ mol$
Ratio: $2.18$, $NaNH_2$ it's the limiting reactant.
Mass of azide:
$0.384\ mol\times 1/3 \times 65.01\ g/mol=8.32\ g$
