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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 Al2(SO4)3 can form under these conditions?

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


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
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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 

Al2(SO4)3 can form under these conditions?

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What scientific concept do you need to know in order to solve this problem?

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.