Answer
The balanced equation for the reaction is:
$N_2(g) + 3I_2(g) --\gt 2NI_3(g)$
Work Step by Step
1. As we have determined in 7.82a, the reactants are: $N_2$ and $I_2$, and the product is : $NI_3$
- Write the unbalanced equation:
$N_2(g) + I_2(g) --\gt NI_3(g)$
2. Balance the number of nitrogen atoms, by putting a "2" as the coefficient of $NI_3$:
$N_2(g) + I_2(g) --\gt 2NI_3(g)$
3. Balance the number of iodine atoms, by putting a "3" as the coefficient of $I_2$:
$N_2(g) + 3I_2(g) --\gt 2NI_3(g)$