Calculate the mass of nitrogen that would be produced when 130 g of sodium azide decomposes

I'm doing a past paper for chemistry and i havnt been taught how to work out questions like these.

Calculate the mass of nitrogen that would be produced when 130 g of sodium azide decomposes?

The mixture of chemicals contains sodium azide (NaN3) which decomposes on heating to form sodium and nitrogen.
2NaN3 -> 2Na + 3N2

Relative atomic masses (Ar): N = 14; Na = 23

Could someone explain how to work out the answer please step by step.

much appreciated.

http://yfrog.com/0xphotoon20110525at0013…

http://yfrog.com/jvphotoon20110525at0013…

:)

This is a question of going from:

grams NaN3 to moles NaN3 to moles N2 to grams N2

To go from grams NaN3 to moles NaN3 you must know the molar mass of NaN3.

molar mass NaN3 = 23.0 g + 3(14.0 g) = 66.0 g/mole
molar mass N2 = 2(14.0 g) = 28.0 g/mole

1st, calculate moles NaN3 = 130 g NaN3(1 mole NaN3)/(66.0 g NaN3) = 1.969 mole NaN3

2nd, calculate moles N2 produced = 1.969 mole NaN3(3 moles N2)/(2 moles NaN3) = 2.954 mole N2
3rd, calculate mass N2 produced = 2.954 mole N2(28.0 g N2)/(1 mole N2) = 82.7 g N2

Hope this is helpful to you. JIL HIR

2NaN3 --> 2Na + 3N2

Given, m(NaN3) = 130g
Molar mass of NaN3 = 23+(14*3) = 65 g/mol
As, n = m/M
n(NaN3) = 130/65 = 2mol

Now, working the number of moles of Nitrogen (N2)

n(N2) = (3/2)*n(NaN3) = (3/2)*2 = 3mol

m(N2) = n*M = 3*(14*2) = 84g