# Question #c5811

Oct 25, 2015

$B a {\left(N {O}_{3}\right)}_{2}$ = $B a O$ + $2 N {O}_{2}$ + $\frac{1}{2} {O}_{2}$

Or

$2 B a {\left(N {O}_{3}\right)}_{2}$ = $2 B a O$ + $4 N {O}_{2}$ + ${O}_{2}$

#### Explanation:

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

$B a {\left(N {O}_{3}\right)}_{2}$ = $B a O$ + $N {O}_{2}$ + ${O}_{2}$

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 $N {O}_{2}$ 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

$B a {\left(N {O}_{3}\right)}_{2}$ = $B a O$ + $2 N {O}_{2}$ + ${O}_{2}$

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 $\frac{1}{2}$) = 6

$B a {\left(N {O}_{3}\right)}_{2}$ = $B a O$ + $2 N {O}_{2}$ + $\frac{1}{2} {O}_{2}$

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.

$\cancel{2}$ [$B a {\left(N {O}_{3}\right)}_{2}$ = $B a O$ + $2 N {O}_{2}$ + $\frac{1}{\cancel{2}} {O}_{2}$]

= $2 B a {\left(N {O}_{3}\right)}_{2}$ = $2 B a O$ + $4 N {O}_{2}$ + ${O}_{2}$