Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 117: 3.67b

15.5g

Molar mass of N$_2$ = 14.0 * 2 = 28.0amu n(N$_2$) = 10.0 / 28.0 = 0.357mole For every 2 moles of NaN$_3$ used there are 3 moles of N$_2$ produced n(NaN$_3$) = 0.357mole * (2 moles of NaN$_3$ / 3 moles of N$_2$) = 0.238mole Molar mass of NaN$_3$ = 23.0 + 3(14.0) = 65.0amu Mass of NaN$_3$ required = 0.238 * 65.0 = 15.5g