John T. answered 04/30/19
Doctoral-level tutoring for STEM, Biostatistics, SAT/ACT, and GRE
70 L of N2 at STP corresponds to 3.125 moles of N2.
3.125 mole of N2 corresponds to 2.083 moles of NaN3 as per stoichiometric coefficients in the balanced reaction.
2.083 moles of NaN3 correspond to 135.42 g NaN3 as per molar mass of 65.0099.