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  • 21-05-2019
  • Chemistry
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How many grams of aluminum hydroxide is in 4.41 mol of Al(OH)^3?

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superman1987
superman1987 superman1987
  • 21-05-2019

Answer:

344.02 g.

Explanation:

  • Knowing that the no. of moles can be calculated using the relation:

no. of moles (n) = mass/molar mass

no. of moles Al(OH)₃ = 4.41 mol & molar mass of Al(OH)₃ = 78.01 g/mol.

∴ mass = no. of moles * molar mass = (4.41 mol)*(78.01 g/mol) = 344.02 g.

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