Answer: 1424 grams
Step-by-step explanation:
![2NaN_3\rightarrow 2a+3N_2](https://img.qammunity.org/2020/formulas/chemistry/college/lfrs0dalz4k8uzbdlunf1sjzwisn81zc42.png)
![Density=(mass)/(Volume)](https://img.qammunity.org/2020/formulas/chemistry/high-school/aysyl7t1ovpe0kyznmirezjgt8r525b796.png)
![1.25g/L=(mass)/(736L)](https://img.qammunity.org/2020/formulas/chemistry/college/h3gv4k85j39cos9nkmafxnaf3vcvun5gt3.png)
![Mass=920g](https://img.qammunity.org/2020/formulas/chemistry/college/drsyjhts1ggrf6m31cbm4abdxdjhps79ke.png)
Thus mass of nitrogen produced is 920 g.
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://img.qammunity.org/2020/formulas/chemistry/high-school/gwh5prgbdt4s2p8o8xquycz897bwt6lvw1.png)
![\text{Number of moles of nitrogen}=(920g)/(28g/mol)=32.8moles](https://img.qammunity.org/2020/formulas/chemistry/college/n9y3f6izuvx3jhdqholznd6gq76vcpdreb.png)
According to stoichiometry:
3 mole of
are produced from= 2 mole of
![NaN_3](https://img.qammunity.org/2020/formulas/chemistry/college/civqahoqjusmii3f4iyhpyo3tucfmkmdj0.png)
Thus 32.8 moles of
are produced from=
of
![NaN_3](https://img.qammunity.org/2020/formulas/chemistry/college/civqahoqjusmii3f4iyhpyo3tucfmkmdj0.png)
Mass of
![NaN_3=moles* {\text {molar mass}}=21.9* 65=1424g](https://img.qammunity.org/2020/formulas/chemistry/college/82xkm875p0lfne0ih3wlijoqhtj0e7sefj.png)
Thus 1424 grams of sodium azide is required to produce 736 L of Nitrogen gas with the density of 1.25 g/L.