Question

Question #c5811

Answer 1

`Ba(NO_3)_2` = `BaO` + `2NO_2` + `1/2O_2`

Or

`2Ba(NO_3)_2` = `2BaO` + `4NO_2` + `O_2`

Explanation:

It would benefit you if you separate every element and tally them up.

`Ba(NO_3)_2` = `BaO` + `NO_2` + `O_2`

Left side:
Ba = 1
N = 2
O = 3 x 2 = 6

Right Side:
Ba = 1
N = 1
O = 1 + 2 + 2 (DO NOT ADD IT ALL UP YET)

Start with the element that is easiest to balance. In this case, the N.

Thus,

Left side:
Ba = 1
N = 2
O = 3 x 2 = 6

Notice that because `NO_2` is a substance, you also have to multiply the O in this substance by 2.

Right Side:
Ba = 1
N = 1 x 2 = 2
O = 1 + (2 x 2) + 2

`Ba(NO_3)_2` = `BaO` + `2NO_2` + `O_2`

Now all that is left to balance is the O. You can this is two ways: one is trial and error, and two, think of a number that can satisfy the equation. In this case, using a fraction is easier because it eliminates the need to increase the number of atoms that are already balanced.

Hence,
Left side:
Ba = 1
N = 2
O = 3 x 2 = 6

Right Side:
Ba = 1
N = 1 x 2 = 2
O = 1 + (2 x 2) + (2 x `1/2`) = 6

`Ba(NO_3)_2` = `BaO` + `2NO_2` + `1/2O_2`

Of course, answer above is already in its reduced form. But if you want whole integers instead of showing fractions, just multiply the WHOLE EQUATION by 2.

`cancel2` [`Ba(NO_3)_2` = `BaO` + `2NO_2` + `1/(cancel 2)O_2`]

= `2Ba(NO_3)_2` = `2BaO` + `4NO_2` + `O_2`