How many moles of al(oh)^3 is in 12.4 g of aluminum hydroxide?

0.1589 mol ≅ 0.16 mol.

Explanation:

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

mass of Al(OH)₃ = 12.4 g & molar mass of Al(OH)₃ = 78.01 g/mol.

∴ n = mass/molar mass = (12.4 g)/(78.01 g/mol) = 0.1589 mol ≅ 0.16 mol.

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