Answer: 136 g of [tex]NaN_3[/tex] must be reacted to inflate an air bag to 70.6 L at STP.
Explanation:
Using ideal gas equation:
[tex]PV=nRT[/tex]
P= pressure of nitrogen gas = 1 atm (at STP)
V =volume of nitrogen gas = 70.6 L
n = number of moles of nitrogen gas = 1 atm (at STP)
R = gas constant = 0.0821 Latm/Kmol
T = temperature of nitrogen gas = 273 K (at STP)
[tex]1atm\times 70.6L=n\times 0.0821 L atm/K mol\times 273[/tex]
[tex]n=3.14[/tex]
For the balanced chemical reaction:
[tex]2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)[/tex]
3 moles of nitrogen are produced by = 2 moles of [tex]NaN_3[/tex]
3.14 moles of nitrogen are produced by =[tex]\frac{2}{3}\times 3.14=2.09[/tex] moles of [tex]NaN_3[/tex]
Mass of [tex]NaN_3[/tex] = Moles × Molar mass = 2.09 mole × 65 g/mol = 136 g