Given unbalanced equation is
"NH"_3(l) + "O"_2(g) -> "NO"(g) + "H"_2"O"(l)NH3(l)+O2(g)→NO(g)+H2O(l)
As a thumb rule "H" and "O"HandO are left for the last. As we see that other than these two atoms there is only one atom, "N"N. That atom is already balanced.
We start with "H" and "O"HandO. 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)xNH3(l)+yO2(g)→xNO(g)+bH2O(l)
To balance hydrogen atoms we multiply ammonia molecule with 2 (set x=2x=2) on reactants side and water molecule by 3 (set b=3b=3) on the product side. Simultaneously multiplying nitric oxide molecule on the product side with 2 We have already set x=2x=2.
The equation looks like
"2NH"_3(l) + y"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l)2NH3(l)+yO2(g)→2NO(g)+3H2O(l)
For the remaining unbalanced element:
Total number of oxygen atoms on products side becomes 2+3=52+3=5. Which gives us number of oxygen molecules as y=5/2y=52 on the reactants side.
"2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l)2NH3(l)+52O2(g)→2NO(g)+3H2O(l)
We know that yy 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))2×(2NH3(l)+52O2(g)→2NO(g)+3H2O(l))
"4NH"_3(l) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(l)4NH3(l)+5O2(g)→4NO(g)+6H2O(l)
