Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)Al2(SO4)3(aq)+6H2O(l) a) Which is the limiting reactant when 0.5molAl(OH)3 and 0.5molH2SO4 are allowed to react? b) How many moles of Al2(SO4)3 can form under these conditions? c) How many moles of the excess reactant remain after the completion of the reaction? (Mw: Al(OH)3=78,H2SO4=98, and Al2(SO4)3=342g/mol)