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

2 Answers
Nov 21, 2015

Answer:

#color(red)(2)NH_3+H_2SO_4->(NH_4)_2SO_4#

Explanation:

The balanced equation is the following:

#color(red)(2)NH_3+H_2SO_4->(NH_4)_2SO_4#

Explanation:
First start by looking at the sulfate group #SO_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 #NH_3# by #color(red)(2)#.

Third, look at the hydrogen atoms, you will find #color(blue)(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

Answer:

#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
#SO_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
#SO_4#: 1 on each side

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