# 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 ###### 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?