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

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

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Answer 1 · 2016-10-30T23:17:43

The balanced equation is $2 {CH}_{3} OH + 3 O_{2} \to 2 {CO}_{2} + 4 H_{2} O$ $2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"$

Explanation:

You follow a systematic procedure to balance the equation.

Start with the unbalanced equation:

${CH}_{3} OH + O_{2} \to {CO}_{2} + H_{2} O$ $"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"$

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

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

The most complicated formula looks like ${CH}_{3} OH$ $"CH"_3"OH"$ . We put a 1 in front of it to remind ourselves that the number is now fixed.

We start with

$1 {CH}_{3} OH + O_{2} \to {CO}_{2} + H_{2} O$ $color(red)(1)"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"$

Balance $C$ $"C"$ :

We have $1 C$ $"1 C"$ on the left, so we need $1 C$ $"1 C"$ on the right. We put a 1 in front of the ${CO}_{2}$ $"CO"_2$ .

$1 {CH}_{3} OH + O_{2} \to 1 {CO}_{2} + H_{2} O$ $color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"$

Balance $H$ $"H"$ :

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

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

$1 {CH}_{3} OH + O_{2} \to 1 {CO}_{2} + 2 H_{2} O$ $color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O"$

Balance $O$ $"O"$ :

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

Uh, oh! Fractions!

We start over, this time doubling all the coefficients.

$2 {CH}_{3} OH + O_{2} \to 2 {CO}_{2} + 4 H_{2} O$ $color(red)(2)"CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"$

Now we can balance $O$ $"O"$ by putting a 3 in front of $O_{2}$ $"O"_2$

$2 {CH}_{3} OH + 3 O_{2} \to 2 {CO}_{2} + 4 H_{2} O$ $color(red)(2)"CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"$

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

Let's check.

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

All atoms balance. The balanced equation is

$2 {CH}_{3} OH + 3 O_{2} \to 2 {CO}_{2} + 4 H_{2} O$ $2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"$

Answer 2 · 2016-10-31T12:06:25

$2 C H_{3} O H$ $2CH_3OH$ $+ 3 O_{2}$ $+ 3O_2$ $\to 2 C O_{2} + 4 H_{2} O$ $rarr 2CO_2 +4H_2O$ .

Explanation:

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

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

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

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