Agustin C. answered 05/07/18
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2 NaN3(s) ——> 2 Na(s) + 3 N2(g)
PV = nRT ——> n=PV/RT
n = 1atm*22L/(0.08206*284K)
n = 0.944 mol N2
MW or MM of NaN3 = 23+14*3 = 65 g/mol
(0.944 mol N2) x (2 mol NaN3 / 3 mol N2) x (65g NaN3 / 1 mol NaN3)
mass of NaN3 = 40.9 g