Balancing Reactions Part 1

Question

____Al   +              ____O₂      →              ____Al₂O₃

_{Please explain how to solve the problem and answer.}

Answer 1

Al + O₂ ==> Al₂O₃

We need to get the same number of each element on both sides of the ==> sign.

As it is, you have 2 Al on the right, but only 1 Al on the left. We fix that as follows:

2 Al + O₂ ==> Al₂O₃

Now we have 3 O on the right and only 2 O on the left. We can fix that by making 6 O on each side....

2 Al + 3 O₂ ==> 2 Al₂O₃ but now we've added more Al to the right. We counter that by increasing Al on the left as follows:

4 Al + 3 O₂ ==> 2 Al₂O₃ and we now have 4 Al on both sides and 6 O on both sides

__4__Al   +             __3__O₂      →              __2__Al₂O₃

Answer 2

Hello, Katrina,

Balancing chemical equations requires some art and patience, but the approach I show here helps build those traits. The thing to remember is that the equation can be literally read as single molecules, made up of individual atoms.

A good way to start is to set up a table to keep track of all the atoms. Leave room in the equation to add a number for the number of atoms of each component under that compound. Then check to see if the sum of all the atoms coming in is equal to the sum of all the atoms leaving as products. There isn't room here for my table, but hopefully you can visualize it.

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In your problem:

____Al   +             ____O₂      →              ____Al₂O₃

Start with the most complex molecule, Al₂O₃. Since we need a whole molecule, let's assign a 1 as it's coefficient. That tells us we need 2 Al atoms coming in, so assign a 2 for Al. But now I'm startled to find that there are three oxygens in the product. Oxygen is diatomic, so it must enter as O₂. I can't assign a whole number to O₂ that will result in an odd number of oxygens. In this case, let's go back and make it two Al₂O₃ molecules. We are now here:

____Al   +             ____O₂      →              _2_Al₂O₃

Then add the coefficients for Al and O₂ that will provide 4 Al and 6 O atoms.

__2_Al   +             _3__O₂      →              __2_Al₂O₃

Double check the result. Two Al atoms on both sides. Six O atoms on both sides. It works!

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These take some practice, so keep at it. One hint: if you have a problem with even/odd numbers like we had above, you can work with fractions as a holding pattern until the equation is balanced. You then multiply them all by the smallest number that would make them whole numbers. In this problem, for example:

____Al   +             ____O₂      →              _1_Al₂O₃ [Note to self: "Darn, I've got an odd number of oxygens in the product. Well, I'm going to try it anyway."]

_1__Al   +             _1/2_O₂      →              _1_Al₂O₃ ["The teacher won't like this, but it works. I'll foot him/her - I'll just multiply everything by 2 to make the 1/2 a whole number."]

_2_Al   +             _1_O₂      →              _2_Al₂O₃ ["This will work, and he./she will never know."]

Have fun,

Bob