Answer
Balanced chemical equation:
$2NH_3(g) + 5F_2(g) --\gt N_2F_4(g) + 6HF(g)$
Work Step by Step
1. Identify the products and reactants, and write the unbalanced equation:
Reactants: $NH_3$ and $F_2$
Products: $N_2F_4$ and $HF$
$NH_3(g) + F_2(g) --\gt N_2F_4(g) + HF(g)$
2. Balance the number o nitrogen atoms, by putting a "2" in the front of $NH_3$.
$2NH_3(g) + F_2(g) --\gt N_2F_4(g) + HF(g)$
3. Balance the number o hydrogen atoms, by putting a "6" in the front of $HF$.
$2NH_3(g) + F_2(g) --\gt N_2F_4(g) + 6HF(g)$
3. There is a total of 10 $F$ atoms on the products side. Balance the number o fluorine atoms, by putting a "5" in the front of $F_2$.
$2NH_3(g) + 5F_2(g) --\gt N_2F_4(g) + 6HF(g)$
