Respuesta :
Answer: 47.2 g of sodium azide
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles of nitrogen}=\frac{\text{Given volume}}{\text {Molar volume}}=\frac{24.4L}{22.4L}=1.09moles[/tex]
The balanced chemical reaction is:
[tex]2NaN_3\rightarrow 2Na+3N_2[/tex]
According to stoichiometry :
3 moles of [tex]N_2[/tex] are produced by = 2 moles of [tex]NaN_3[/tex]
Thus 1.09 moles of [tex]N_2[/tex] are produced by =[tex]\frac{2}{3}\times 1.09=0.73moles[/tex] of [tex]NaN_3[/tex]
Mass of [tex]NaN_3=moles\times {\text {Molar mass}}=0.73moles\times 65g/mol=47.2g[/tex]
Thus 47.2 g of sodium azide are required to produce 24.4 L of nitrogen gas at standard temperature and pressure
