Question #de9da

1 Answer
Oct 4, 2015

#"CH"_3"COOH"_text((l]) + 2"O"_text(2(g]) -> 2"CO"_text(2(g]) + 2"H"_2"O"_text((g])#


You can balance this chemical equation by inspection.

Start by taking a look at the unbalanced equation

#"CH"""_3"COOH"_text((l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((g])#

How many atoms of carbon do you have on the reactants' side? Two, both of the as a part of acetic acid, #"CH"""_3"COOH"#.

How many carbon atoms to you have on the products' side? One, which is a part of the carbon dioxide, #"CO"_2#. In order for the Law of mass conservation to be obeyed, you need to have equal numbers of atoms of each element on both sides of the equation.

This means that you need to multiply the carbon dioxide by #color(blue)(2)# to get th number of atoms of carbon to match.

#"CH"""_3"COOH"_text((l]) + "O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g]) + "H"_2"O"_text((g])#

What about the hydrogen atoms? You have four hydrogen atoms on the reactants' side, all of them as a part of the acetic acid, and only two on the products' side, both part of the water.

Once again, multiply by #color(red)(2)# the water to balance the hydrogen atoms out.

#"CH"""_3"COOH"_text((l]) + "O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g]) + color(red)(2)"H"_2"O"_text((g])#

Finally, look at the oxygen atoms. You have four on the reactants' side, two in acetic acid and two in molecular oxygen, and six on the products' side, four coming from the carbon dioxide and two coming from the water.

To balance these atoms out, multiply the molecular oxygen by #color(green)(2)# to get

#"CH"""_3"COOH"_text((l]) + color(green)(2)"O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g]) + color(red)(2)"H"_2"O"_text((g])#

And now the chemical equation is balanced.

As far as the physical states go, your unbalanced equation included them as well. Those little letters that are added as subscripts at the end of a chemcial formula represent the physical state of the compound in a reaction.

In your case, you have liquid acetic acid, symbolized by #"(l)"#, reacting with oxygen in the gaseous state, symbolized by #"(g)"#.

The products will be carbon dioxide in the gaseous state, #"(g)"#, and water in the gaseous state, #"(g")#.