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Question:

Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)→Al2(SO4)3(aq)+6H2O(l)?

Which reagent is the limiting reactant when 0.650mol Al(OH)3 and 0.650mol H2SO4 are allowed to react?How many moles of Al2(SO4)3 can form under these conditions?How many moles of the excess reactant remain after the completion of the reaction?Please and thank you

Answer:

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0.65 mol of aluminium hydroxide requires 0.65 x 1.5 mol of H2SO4 to react completelySo there aren't enough moles of H2SO4, and so there will be surplus Al(OH)3Only 0.65/3 mol of Al2(SO4)3 can form, and 2/3 of the Al(OH)3 will react, leaving one third unreacted.

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