How do you balance out this equation?

Question

1. ammonia +oxygen ® nitrogen +water

2. ethane +oxygen ® carbon dioxide +water

Answer 1

You must convert the words to chemical formulae first of all. Simple balancing of the resulting equations is then just a matter of making sure that you have the same number of each kind of atoms one both sides of the equation.

The first one becomes

NH3  + O2  --->  N2  +  H2O

There are 3 H on the left in NH3, but only 2 H on the right in H2O, and only 1 N in NH3 but 2 N in N2. So if we change the equation to

2 NH3 + O2  --->  N2 + 3 H2O

we get 2 N, 6 H, and 2 O atoms on both the left and the right sides of the equation, and we can say that the equation is balanced.

The second one becomes

C2H6 + O2  --->  CO2 + H2O

Use what you learned from the first equation to solve this one. You know you must have 2 CO2 on the right and you will need more than 1 O2 on the left...

Answer 2

You need to write the formulas for the different molecules, and then make sure that you have the same number of atoms on both sides.

1. ?NH3 + ?O2 --> ?N2 + ?H2O

To get N2, you are going to need 2 NH3 (why?) So

2NH3 + ?O2 --> N2 + ?H2O

Now you have 6 H atoms in your reactants, so you need 6 H atoms in your products:

2NH3 + ?O2 --> N2 + 3H2O

Now balance oxygens:

2NH3 + 3/2 O2 --> N2 + 3H2O

But people don't like half molecules, so multiply through to get rid of the fraction:

4NH3 + 3 O2 --> 2 N2 + 6 H2O

The second problem is very similar.  Over to you.

Answer 3

Hi :) The best way to tackle these is look for the element that only appears once on each side, or is lonesome and not in a compound with another element and balance that out first, then simply balance the others to match it!

1) 4 NH3 + 3 O2 → 2 N2 + 6 H2O

2) 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

For example in the first one, you start with:

NH3 + O2 → N2 + H20

There is N2 on it's own so we know there has to be at least 2 ammonias. This means there are 6 Hydrogens so you need at least 3 H2O, and therefore 1.5 oxygens. 1.5 isn't an ideal number to have, so to make it an integer, double all of the prefixes :).

Not too bad really :)