Aluminum hydroxide reacts with sulfuric acid as follows:2 Al(OH)3(s) + 3 H2SO4(aq) Al2(SO4)3(aq) + 6 H2O(l)Which is the limiting reactant when 0.500 mol Al(OH)3 and 0.500 mol H2SO4 are allowed to react? How many moles of Al2(SO4)3 can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

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periwinklecat295Lv1

3 Apr 2020

Aluminum hydroxide reacts with sulfuric acid as follows:
2 Al(OH)₃(s) + 3 H₂SO₄(aq) Al₂(SO₄)₃(aq) + 6 H₂O(l)
Which is the limiting reactant when 0.500 mol Al(OH)₃ and 0.500 mol H₂SO₄ are allowed to react? How many moles of Al₂(SO₄)₃ can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

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Deanna HettingerLv2

28 May 2020

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Related questions

Aluminum hydroxide reacts with sulfuric acid as follows:

2Al(OH)3(s)+3H2SO4(aq)âAl2(SO4)3(aq)+6H2O(l)

1) Which reagent is the limiting reactant when 0.650 mol Al(OH)3 and 0.650 mol H2SO4 are allowed to react?

2) How many moles of Al2(SO4)3 can form under these conditions? Express the amount in moles to three significant digits.

3) How many moles of the excess reactant remain after the completion of the reaction? Express the amount in moles to three significant digits.

Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)âAl2(SO4)3(aq)+6H2O(l) How many moles of Al2(SO4)3 can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

2 Al(OH)3(s) + 3 H2SO4(aq) --> Al2(SO4)3(aq) + 6 H2O(l) If 0.500 mol Al(OH)3 and 0.500 mol H2SO4 react, and 48.2 g Al2(SO4)3 are collected from the experiment, what is the percent yield? The molar mass of Al2(SO4)3 is 342.17 g/mol. Be sure to include "%" units with your answer.