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"#

To balance the equation, you adhere to a methodical process.

To begin, consider the imbalanced 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"#.

Starting with the formula that appears to be the most complex is another helpful approach.

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 begin 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.

Oh no! Fractions!

Restarting, we double all the coefficients this time.

#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"#

Now that each formula has a coefficient, the equation ought to be balanced.

Let's investigate.

#"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#

Atoms all balance. The equation that is in balance is

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

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

Verify now that there are the same number of individual atoms in L.H.S. and R.H.S.

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Answer 3

The balanced chemical equation for the combustion of methanol (CH3OH) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:

2CH3OH + 3O2 → 2CO2 + 4H2O

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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