How many grams of sodium azide(NaN3) are required to produce 19.00 L of N2 at 293K and 775 mm HG?

Question

2NaN3(s)-->2Na(s) 3N2(g) please explain the steps thank you so much

Answer 1

first we find the moles of N2 they wish to be produced:

PV=nRT

(775Torr)(19.00L) = n (62.36 L-Torr/mol-K)(293K)

n= 0.8059 moles of N2

the equation says that 2 moles NaN3(s)--> 3 moles N2(g), so, find the moles of NaN3 needed:

0.8059 moles N2 @ 2 mol NaN3 / 3 mol N2 = 0.5373 moles NaN3

now use molar mass to get grams of NaN3:

0.5373 moles of NaN3 @ 65.01 g/mol = 34.93 grams NaN3

your answer (3sigfigs) : = 34.9 grams of NaN3