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 H2OWhen 45.0 g of H2SO4 and 25.0 g of Al(OH)3 are reacted, how many grams of Al2(SO4)3 are produced? Which reactant is in excess?

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(molar mass Al2(SO4)3)→ mass 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

When 45.0 g of H2SO4 and 25.0 g of Al(OH)3 are reacted, how many grams of Al2(SO4)3 are produced? Which reactant is in excess?

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