## General Chemistry: Principles and Modern Applications (10th Edition)

Using ideal gas law: $n=\frac{PV}{RT}$ $=\frac{(776\, mmHg\times\frac{1\,atm}{760\, mmHg})(20.0\,L)}{(0.08206\,L\,atm\,K^{-1}mol^{-1})(30.0+273)K}$ $=0.8213\,mol$ Grams of $NaN_{3}=0.8213\,mol\times\frac{65.0\,g\,NaN_{3}}{1\,mol}$ $=53.4\,g\,NaN_{3}$