Respuesta :
Answer: 1424 grams
Explanation:
[tex]2NaN_3\rightarrow 2a+3N_2[/tex]
[tex]Density=\frac{mass}{Volume}[/tex]
[tex]1.25g/L=\frac{mass}{736L}[/tex]
[tex]Mass=920g[/tex]
Thus mass of nitrogen produced is 920 g.
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of nitrogen}=\frac{920g}{28g/mol}=32.8moles[/tex]
According to stoichiometry:
3 mole of [tex]N_2[/tex] are produced from= 2 mole of [tex]NaN_3[/tex]
Thus 32.8 moles of [tex]N_2[/tex] are produced from=[tex]\frac{2}{3}\times 32.8=21.9moles[/tex] of [tex]NaN_3[/tex]
Mass of [tex]NaN_3=moles\times {\text {molar mass}}=21.9\times 65=1424g[/tex]
Thus 1424 grams of sodium azide is required to produce 736 L of Nitrogen gas with the density of 1.25 g/L.
