What mass of NaN3(s) must be reacted to inflate an air bag to 65.5 L at STP?-Chem Webassign?
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1 mole of N2 at STP = 22.4 L
65.5 L / 22.4 L/mole = 2.92 moles of N2
2 moles of NaN3 will produce 3 moles of N2
(2/3) x 2.92 = 1.95 moles of NaN3 required
1.95 moles x 65.01 g/mole = 126.7 g (127 g)