Aluminum hydroxide reacts with sulfuric acid as follows:
2 Al(OH)3 (s) + 3 H2SO4 (aq) ---> Al2(SO4)3 (aq) + 6 H2O (l)
(a) Which reagent is the limiting reactant when 0.450 mol Al(OH)3 and 0.550 mol H2SO4 are allowed to react? How many moles of the excess reactant remain after the completion of the reaction? Please explain in details.
0.450 mol of Al(OH)3 requires, 0.450 mol Al(OH)3 x [3 mol H2SO4 / 2 mol Al(OH)3] = 0.675 ...
The expert examines aluminum hydroxide and sulfuric acid reactions.