Ionic compounds dissolve in water, and their ions separate from each other. If a sample of Al2S3 containing 16 aluminum ions is dissolved in water, how many sulfide ions will be present?
So, two cations and three anions.
Would we have one sulfide ion left? Or am I going about this incorrectly?
Yes, you missed the boat. A simplified dissociation is shown as
Al2S3 ==> 2Al^2+ + 3S^2-
So if you have 16 Al ions you must have 24 sulfide ions.
16 Al ions x (3 sulfide ions/2 aluminum ions) = 24 sulfide ions.
Technically you have a problem with this since Al2S3 dissolves in water this way as Al2S3 + 6HOH --> 2Al(OH)3 + 3H2S and since neither Al(OH)3 nor H2S in water gives that ratio you don't get that many Al ions nor that many sulfide ions. [The Ksp for Al(OH)3 is extremely small and the k1 and k2 for H2S is very small]. But the first answer I gave surely is the correct answer to your problem; i.e., that's obviously how the author meant for it to be interpreted.