Answer
The balanced equation for the reaction is:
$2NI_3(g) --\gt N_2(g) + 3 I_2(g)$
Work Step by Step
1. As we have determined in 7.81a, the reactant is: $NI_3$, and the products are: $N_2$ and $I_2$.
- Write the unbalanced equation:
$NI_3(g) --\gt N_2(g) + I_2(g)$
2. Balance the number of nitrogen atoms, by putting a "2" as the coefficient of $NI_3$.
$2NI_3(g) --\gt N_2(g) + I_2(g)$
3. Balance the number of iodine atoms, by putting a "3" as the coefficient of $I_2$.
$2NI_3(g) --\gt N_2(g) + 3 I_2(g)$
- The equation is balanced.