Question

Question #84161

Answer 1

Think about balancing the positive charge of the cation and the negative charge of the anion.

You get a +3 charge from the aluminium ion, `Al^(color(red)(3+))`, and a -2 charge from the sulfate ion,`SO_4^(color(blue)(-2))`. Notice that you can use 2 aluminium ions, for a total positive charge of +6, and 3 sulfate ions, for a total negative charge of -6, and achieve neutrality.

So, you need `2Al^(color(red)(3+))` and `3SO_4^(color(blue)(2-))`, which will make your formula

`Al_(color(blue)(2))(SO_4)_(color(red)(3))`

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The subscripts are equal to the original charges of the ions. This comes about from the cross-over rule for ionic compounds, which you can read about here:

http://socratic.org/chemistry/ionic-bonds-and-formulas/writing-ionic-formulas/cross-over-rule-for-ionic-formulas