A general formula is:
#"metal oxide"# #+ "acid"->"water" + "salt"#
So, the metal oxide is #Al_2O_3#, the acid is #H_2SO_4# and the salt is therefore #Al_2(SO_4)_3#
The unbalanced reaction is therefore #Al_2O_3+H_2SO_4->Al_2(SO_4)_3+H_2O#
The reaction needs three #SO_4^(2-)# ions at the end, so 3 #H_2SO_4#s are needed which means three waters are produced, giving:
#Al_2O_3+3H_2SO_4->Al_2(SO_4)_3+3H_2O#
However, we need the ionic equation, so we first split the compounds into their ions:
#Al_2O_3+3H^(+)+3HSO_4^(-)->2Al^(3+)+3(SO_4)^(2-)+3H^(+)+3OH^-#
Finally we remove similar ions
#Al_2O_3+3HSO_4^(-)->2Al^(3+)+3(SO_4)^(2-)+3OH^-#
However, there is one slight problem. During the reaction, there will be spare #HSO_4^-# ions, which are acidic, and #OH^-# which is basic. So, these will react accordingly:
#OH^(-)+HSO_4^(-)->SO_4^(2-)+H_2O#
Bringing us back to this but it has been changed (however the other hydrogen atom has dissociated due to the above reaction):
#Al_2O_3+6H^(+)+3SO_4^(2-)->2Al^(3+)+3(SO_4)^(2-)+3H_2O#
