Question

Is the following equation balanced: NH2CH2COOH (s) + 9/4 O2 --> 2 CO2 (s) + 5/2 H2O (l) + 1/2 N2 (g)?

Answer 1

No, balanced equation should be

`2NH_2CH_2COOH` + `9/2O_2` `rarr` `4CO_2` + `5H_2O` + `N_2`

Explanation:

Let's write first the 'original' equation:

`NH_2CH_2COOH` + `O_2` `rarr` `CO_2` + `H_2O` + `N_2` (unbalanced)

Let's begin tallying the atoms based on their respective subscripts

left side:

N = 1
H = 5
C = 2
O = 2 + 2 (do not add this up yet)

right side:

N = 2
H = 2
C = 1
O = 2 + 1 (do not add this up yet)

Since you have a rather large molecule on the left and simple molecules on the right, you have to choose one of those simple molecules to start balancing.

I'll start with `H_2O`. Since this is a substance, no matter what coefficient you multiply to `H` atom, you also have to apply to its bonded `O` atom.

`NH_2CH_2COOH` + `O_2` `rarr` `CO_2` + `color (red) 5H_2O` + `N_2`

left side:

N = 1
H = 5
C = 2
O = 2 + 2

right side:

N = 2
H = 2 x `color (red) 5` = 10
C = 1
O = 2 + (1 x `color (red) 5`)

Notice that you now have 10 `H` atoms on your right, so you need to also have 10 `H` atoms on the left.

`color (green) 2NH_2CH_2COOH` + `O_2` `rarr` `CO_2` + `5H_2O` + `N_2`

left side:

N = 1 x `color (green) 2` = 2
H = 5 x `color (green) 2` = 10
C = 2 x `color (green) 2` = 4
O = (2 x `color (green) 2`) + 2

right side:

N = 2
H = 2 x 5 = 10
C = 1
O = 2 + (1 x 5)

Again, since your `H` atom is bonded to the `N`, `C` and `O` atoms, you also need to multiply the coefficient to them. Now let's balance the `C` atom.

`2NH_2CH_2COOH` + `O_2` `rarr` `color (blue) 4CO_2` + `5H_2O` + `N_2`

left side:

N = 1 x 2 = 2
H = 5 x 2 = 10
C = 2 x 2 = 4
O = (2 x 2) + 2

right side:

N = 2
H = 2 x 5 = 10
C = 1 x `color (blue) 4` = 4
O = (2 x `color (blue) 4`) + (1 x 5) = 13

Now the only atom left to balance is the `O` atom. Since the total number of `O` atoms on the right side is an odd number, I can use my knowledge on fractions to balance the equation.

`2NH_2CH_2COOH` + `color (magenta) (9/2)O_2` `rarr` `color (blue) 4CO_2` + `5H_2O` + `N_2`

left side:

N = 1 x 2 = 2
H = 5 x 2 = 10
C = 2 x 2 = 4
O = (2 x 2) + (2 x `color (magenta) (9/2)`) = 13

right side:

N = 2
H = 2 x 5 = 10
C = 1 x 4 = 4
O = (2 x 4) + (1 x 5) = 13

The equation is now balanced.