Question #c5811

1 Answer
Oct 25, 2015

Answer:

#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#