Limiting Reagent Video Lessons

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Problem: Aluminum sulfate has many uses in industry. It can be prepared by reacting aluminum hydroxide and sulfuric acid: 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O(a) When 35.0 g of Al(OH)3 and 35.0 g of H2SO4 are reacted, how many moles of Al2(SO4)3 can be produced? Which substance is the limiting reactant?

FREE Expert Solution

Calculate Al2(SO4)3 from Al(OH)3 :


Mass Al(OH)3 (molar mass Al(OH)3)→ Moles of Al(OH)3(mole-to-mole comparison)  Moles of Al2(SO4)3


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Problem Details

Aluminum sulfate has many uses in industry. It can be prepared by reacting aluminum hydroxide and sulfuric acid:

 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

(a) When 35.0 g of Al(OH)3 and 35.0 g of H2SO4 are reacted, how many moles of Al2(SO4)3 can be produced? Which substance is the limiting reactant?

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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. You can view video lessons to learn Limiting Reagent. Or if you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.