How do you balance CH3OH + O2 — CO2 + H2O? | Socratic

Or you want a quick look:

You follow a systematic procedure to balance the equation.

Start with the unbalanced equation:

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#"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"#

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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.

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#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"#

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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"#

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