Question

How do you balance CH3OH + O2 -- CO2 + H2O?

Answer 1

The balanced equation is `2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"`

Explanation:

You follow a systematic procedure to balance the equation.

Start with the unbalanced equation:

`"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"`

A method that often works is first to balance everything other than `"O"` and `"H"`, then balance `"O"`, and finally balance `"H"`.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like `"CH"_3"OH"`. We put a 1 in front of it to remind ourselves that the number is now fixed.

We start with

`color(red)(1)"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"`

Balance `"C"`:

We have `"1 C"` on the left, so we need `"1 C"` on the right. We put a 1 in front of the `"CO"_2`.

`color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"`

Balance `"H"`:

We can't balance `"O"` because we have two oxygen-containing molecules without coefficients. ∴ Let's balance `"H"` instead.

We have `"4 H"` on the left, so we need `"4 H"` on the right. There are already `"2 H"` atoms on the right. We must put a 2 in front of the `"H"_2"O"`.

`color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O"`

Balance `"O"`:

We have fixed `"4 O"` on the right and `"1 O"` on the left. We need `"3 O"` on the left.

Uh, oh! Fractions!

We start over, this time doubling all the coefficients.

`color(red)(2)"CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"`

Now we can balance `"O"` by putting a 3 in front of `"O"_2`

`color(red)(2)"CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"`

Every formula now has a coefficient. We should have a balanced equation.

Let's check.

`"Atom" color(white)(m)"lhs"color(white)(m)"rhs"`
`color(white)(m)"C"color(white)(mml)2color(white)(mm)2`
`color(white)(m)"H"color(white)(mml)8color(white)(mm)8`
`color(white)(m)"O"color(white)(mml)8color(white)(mm)8`

All atoms balance. The balanced equation is

`2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"`

Answer 2

`2CH_3OH` `+ 3O_2` `rarr 2CO_2 +4H_2O` .

Explanation:

• First try to balance atoms other than O and H. So first balance the C atoms.
  `2CH_3OH` `+ O_2` `rarr 2CO_2 +H_2O` .
• Then balance the H atoms, as no. of H atoms is less than no. of O atoms.
  `2CH_3OH` `+ O_2` `rarr 2CO_2 +4H_2O` .
• Now, balance the remaining O atoms.
  `2CH_3OH` `+3 O_2` `rarr 2CO_2 +4H_2O` .

Now, check that the no. of all individual atoms in L.H.S and R.H.S are equal.