# How do you balance NH_3(l) + O_2(g) -> NO(g) + H_2O(l)?

Mar 24, 2016

$\text{4NH"_3(l) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O} \left(l\right)$

#### Explanation:

Given unbalanced equation is

$\text{NH"_3(l) + "O"_2(g) -> "NO"(g) + "H"_2"O} \left(l\right)$
As a thumb rule $\text{H" and "O}$ are left for the last. As we see that other than these two atoms there is only one atom, $\text{N}$. That atom is already balanced.

We start with $\text{H" and "O}$. 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)
$x \text{NH"_3(l) + y"O"_2(g) -> x"NO"(g) + b"H"_2"O} \left(l\right)$

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

$\text{2NH"_3(l) + y"O"_2(g) -> 2"NO"(g) + 3"H"_2"O} \left(l\right)$

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 $y = \frac{5}{2}$ on the reactants side.

$\text{2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O} \left(l\right)$
We know that $y$ needs to be whole number. Therefore multiply the equation with 2 to obtain balanced equation.

$2 \times \left(\text{2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O} \left(l\right)\right)$
$\text{4NH"_3(l) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O} \left(l\right)$