Question

How many grams of  NaN3(s) would be decomposed by 125 kJ of heat?

Answer 1

The enthalpy of reaction is opposite to the enthalpy of formation of NaN3:

ΔH° =-21300 J/mol =-21.3 kJ/mol

$$n(NaN3)=125/21.3=5.87 mol$$

M(NaN3)=65 g/mol

$$m(NaN3)=nM=5.87*65=381.5 g$$