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:

0.183 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?