## Chemistry and Chemical Reactivity (9th Edition)

$8.32\ g$
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: $0.384\ mol\times 1/3 \times 65.01\ g/mol=8.32\ g$