Which coefficients balance this equation? #?Al + ?O_2 -> ?Al_2O_3#

You should get:

#stackrel("Reactants Side")overbrace(color(red)(4)"Al"(s) + color(red)(3)"O"_2(g)) -> stackrel("Products Side")overbrace(color(red)(2)"Al"_2"O"_3(s))#

To balance the reaction, simply add whole-number coefficients to make sure the number of each element on each side is equal. If you have prime number quantities of an element, balance those first to save time/hassle.

You started with:

#stackrel("Reactants Side")overbrace("Al"(s) + "O"_2(g)) -> stackrel("Products Side")overbrace("Al"_color(red)(2)"O"_color(red)(3)(s))#

Since you have #2# aluminum atoms on the products side, you need #2# on the reactants side, and you only have #1#. But, there's a problem:

• You have #2# oxygen atoms on the reactants side and #3# oxygen atoms on the products side.
• #2# and #3# are both prime numbers, so you cannot add whole-number coefficients on only one side---you need to add them on both sides.

So, you'll need to balance the oxygens first if you want to do this problem in as few steps as possible. For this, balance by finding the common multiple. #2xx3 = 3xx2 = 6#, so your first step is:

#stackrel("Reactants Side")overbrace("Al"(s) + color(red)(3)"O"_2(g)) -> stackrel("Products Side")overbrace(color(red)(2)"Al"_2"O"_3(s))#

Finally, you have achieved #2xx2 = 4# aluminum atoms on the products side and you only have #1# on the reactants side, so multiply #"Al"# by #4# to finish.

#stackrel("Reactants Side")overbrace(\mathbf(color(blue)(color(red)(4)"Al"(s) + color(red)(3)"O"_2(g)))) -> stackrel("Products Side")overbrace(\mathbf(color(blue)(color(red)(2)"Al"_2"O"_3(s))))#

You know you are done because you have:

• #4xx"Al"# vs. #2xx2xx"Al"#
• #3xx2xx"O"# vs. #2xx3xx"O"#