Given unbalanced equation is

#"NH"_3(l) + "O"_2(g) -> "NO"(g) + "H"_2"O"(l)#

As a thumb rule #"H" and "O"# are left for the last. As we see that other than these two atoms there is only one atom, #"N"#. That atom is already balanced.

We start with #"H" and "O"#. We observe that out of these two, hydrogen has more number of atoms in the equation. So we take it first.

Let the balanced equation be: (notice that number of N atoms have been kept balanced)

#x"NH"_3(l) + y"O"_2(g) -> x"NO"(g) + b"H"_2"O"(l)#

To balance hydrogen atoms we multiply ammonia molecule with 2 (set #x=2#) on reactants side and water molecule by 3 (set #b=3#) on the product side. Simultaneously multiplying nitric oxide molecule on the product side with 2 We have already set #x=2#.

The equation looks like

#"2NH"_3(l) + y"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l)#

For the remaining unbalanced element:

Total number of oxygen atoms on products side becomes #2+3=5#. Which gives us number of oxygen molecules as #y=5/2# on the reactants side.

#"2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l)#

We know that #y# needs to be whole number. Therefore multiply the equation with 2 to obtain balanced equation.

#2xx("2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l))#

#"4NH"_3(l) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(l)#