Answer:
47.2 g of NaN3.
Step-by-step explanation:
Equation of the reaction
2NaN3 --> 2Na + 3N2(g)
Molar mass of N2 = (14*2)
= 28 g/mol
Number of moles = mass*molar mass
= 30.5/28
= 1.09 moles of N2
Using stoichiometry, 2 moles of sodium azide decomposes to give 3 moles of nitrogen.
1.09 moles of N2 will give,
= 2/3 * 1.09
= 0.726 moles of NaN3
Molar mass of NaN3 = 23 + (14*3)
= 65 g/mol
Mass = number of moles * molar mass
= 65 * 0.726
= 47.2 g of NaN3