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Problem: You may want to reference (Pages 106 - 109)Section 3.7 while completing this problem.Aluminum hydroxide reacts with sulfuric acid as follows: 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)0.550 moles Al(OH)3 and 0.550 moles H2SO4 are allowed to react. How many moles of the excess reactant remain after the completion of the reaction?

FREE Expert Solution

Find the limiting reactant, which is the reactant that forms the less amount of product

  • determines the maximum amount of the product
  • limiting reactant → gets used up once reaction is complete


Balanced equation: 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)


Mole to mole comparison:

  • 1 mol Al(OH)3 forms 1 mol Al2(SO4)3
  • 3 moles H2SOforms 1 mol Al2(SO4)3


Since moles of  Al(OH)3  and H2SO4 are equal  limiting reactant = H2SO4

  • excess reactant = Al(OH)3


The moles of Al2(SO4)3 formed under the given conditions is:


0.550 mol H2SO4×1 mol Al2(SO4)33 mol H2SO4=0.183 mol Al2(SO4)3


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Problem Details

You may want to reference (Pages 106 - 109)Section 3.7 while completing this problem.

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

0.550 moles Al(OH)3 and 0.550 moles H2SO4 are allowed to react. How many moles of the excess reactant remain after the completion of the reaction?

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Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.