Answer
4N$H_{3}$(g) +5 $O_{2}$(g) $\longrightarrow$ 4NO (g) +6$H_{2}$O(aq)
This is the complete balanced reaction with all proper symbols.
Work Step by Step
N$H_{3}$(g) + $O_{2}$(g) $\longrightarrow$ NO (g) +$H_{2}$O(aq)
This is the chemical reaction with the proper state symbols.
In this reaction oxygen gas enters as a pure substance, and it forms two substances NO and $H_{2}$O .
To balance the chemical reaction add 4 before N$H_{3}$ and before NO.
4N$H_{3}$(g) + $O_{2}$(g) $\longrightarrow$ 4NO (g) +$H_{2}$O(aq)
As there are 12 atoms H on the left and 2 on the right, add 6 before $H_{2}$O.
4N$H_{3}$(g) + $O_{2}$(g) $\longrightarrow$ 4NO (g) +6$H_{2}$O(aq)
There are 10 atoms O on the right (4+6), so add 5 before $O_{2}$
4N$H_{3}$(g) +5 $O_{2}$(g) $\longrightarrow$ 4NO (g) +6$H_{2}$O(aq)
This is the complete balanced reaction with all proper symbols.
