Question

How many moles of Al(OH)^3 is in 12.4 g of aluminum hydroxide?

Answer 1

0.1589 mol ≅ 0.16 mol.

Step-by-step explanation:

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

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.