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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 H2SO4 forms 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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