Question

How many grams of aluminum hydroxide is in 4.41 mol of Al(OH)^3?

Answer 1

344.02 g.

Step-by-step 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.