Limiting Reactants (Section) Aluminum hydroxide reacts with sulfuric acid as follows: Which is the limiting reactant when 0.500 mol AI(OH)3 and 0.500 mol H2SO4 are allowed to react? How many moles of AI2(SO4)3 can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?
Solution 78 Step 1 of 3 Step 2 of 3 Moles of Al2(SO4)3 formed Moles of Al2(SO4)3= 1 mol Al2(SO4)3/3 mol H2SO4*0.500mol H2SO4 =0.1667 mol Al2(SO4)3. =0.167mol.