# How do you balance NH_3 + H_2SO_4 -> (NH_4)_2SO_4?

Nov 21, 2015

$\textcolor{red}{2} N {H}_{3} + {H}_{2} S {O}_{4} \to {\left(N {H}_{4}\right)}_{2} S {O}_{4}$

#### Explanation:

The balanced equation is the following:

$\textcolor{red}{2} N {H}_{3} + {H}_{2} S {O}_{4} \to {\left(N {H}_{4}\right)}_{2} S {O}_{4}$

Explanation:
First start by looking at the sulfate group $S {O}_{4}$, you will need to leave this group intact.

Second, we have two nitrogen atoms in the product side, and only one in the reactants side, therefore, we can multiply $N {H}_{3}$ by $\textcolor{red}{2}$.

Third, look at the hydrogen atoms, you will find $\textcolor{b l u e}{8}$ in each side of the reaction, and therefore, your reaction is balanced.

In some cases, I prefer to balance equations using groups (or polyatomic ions) rather than atoms.

Nov 21, 2015

2NH_3 + H_2SO_4 → (NH_4)_2SO_4

#### Explanation:

You need to make sure that all the elements are equal on both sides.

The first step is write down what you have

NH_3 + H_2SO_4 → (NH_4)_2SO_4

N: 1 the side in front of the arrow, 2 on the side after the arrow
H: 5 on the side before the arrow, 8 on the side after the arrow
$S {O}_{4}$: 1 on each side

Now you can see that you need to change the amount of N and H on the side in front of the arrow, to match the amount of N and H after the arrow.

Step two add a number in front of the elements to make both side match:

2NH_3 + H_2SO_4 → (NH_4)_2SO_4

Step three, check again, make changes if necessary:
N: 2 on both sides
H: 8 on both sides
$S {O}_{4}$: 1 on each side

As you can see, the reaction is now balanced :)