# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179g: 85

$8.32\ g$

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

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