How do you balance __ NH3 + __O2--> __NO2 + __H2O? The number in the parenthasis are subscribts, thank you all so much!!!?

2 Answers

Answer:

#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#

Explanation:

There is no fixed method actually. You have to take a random number as the number of molecules for #NH_3#, then calculate result.

As for here, I have taken #4# as the number of molecules of #NH_3#. That means #4# nitrogen atoms and #12# hydrogen atoms are taking part in this reaction.

Now, hydrogen atoms are present only in #H_2O# as a product in this reaction. So that makes #6# molecules of #H_2O# (as there are #12# hydrogen atoms).

Now, when we calculated the number of molecules of #H_2O#, we also got the number of oxygen atoms present there.

On the other hand, as there are #4# molecules of #NH_3#, there has to be #4# molecules of #NO_2# as well to balance the number of nitrogen atoms in both side of the reaction.

That makes #8# another atoms of oxygen.

So now the total number of oxygen atoms is #(8+6)# or #14#, which means #14# atoms or #7# molecules of oxygen.

So the final balanced reaction is :

#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#

Oct 24, 2015

Answer:

Two possible answers: (1) #2NH_3# + #7/2O_2# = #2NO_2# + #3H_2O# or (2) #4NH_3# + #7O_2# = #4NO_2# + #6H_2O#

Explanation:

First, you need to tally all the atoms.

#NH_3# + #O_2# = #NO_2# + #H_2O#

Left side:
N = 1
H = 3
O = 2

Right side:
N = 1
H = 2
O = 2 + 1 (two from the #NO_2# and one from #H_2O# ; DO NOT ADD IT UP YET )

You always have to find the simplest element that you can balance (in this case, the H).

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

Right side:
N = 1
H = 2 x 3 = 6
O = 2 + (1 x 3) from the O in #H_2O#

Notice that with balancing the atoms, you do not forget that they are part of a substance - meaning, you have to multiply everything.

#2NH_3# + #O_2# = #NO_2# + #3H_2O#

Now, the N is not balanced so you need to multiply the right side N by 2.

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

Right side:
N = 1 x 2 = 2
H = 2 x 3 = 6
O = (2 x 2) + (1 x 3) = 7

#2NH_3# + #O_2# = #2NO_2# + #3H_2O#

Now, the only element left to be balanced is O. Since 7 is an odd number, you can use a fraction for the equation to be in its reduced form.

Left side:
N = 1 x 2 = 2
H = 3 x 2 = 6
O = 2 x 3.5 = 7 (the decimal 3.5 can be written as #7/2#)

Right side:
N = 1 x 2 = 2
H = 2 x 3 = 6
O = (2 x 2) + (1 x 3) = 7

Hence,

#2NH_3# + #7/2O_2# = #2NO_2# + #3H_2O#

but if you want the equation to show whole numbers only, you can always multiply everything by the denominator in the fraction.

#(cancel 2)# [#2NH_3# + #7/(cancel 2)O_2# = #2NO_2# + #3H_2O#]

=

#4NH_3# + #7O_2# = #4NO_2# + #6H_2O#