Question

Calculate the number of milliliters of 0.664 M Ba(OH)2 required to precipitate all of the Al3+ ions in 154 mL of 0.408 M AlCl3 solution as Al(OH)3. The equation for the reaction is: 2AlCl3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) + 3BaCl2(aq) _____mL Ba(OH)2

Answer 1

Balanced chemical equation is

2AlCl_3(aq) + 3Ba(OH)_2(aq) → 2Al(OH)_3(s) + 3BaCl_2(aq)

No. of moles = molarity X volume of solution in liter

moalrity of AlCl₃ = 0.408 M

volume of AlCl₃ = 0.154 liter (1 ml = .001 L then 154 ml = 154 X 0.001 = 0.154 L)

moles of AlCl₃ = 0.408 X 0.154 = 0.062832 moles

According to balanced reaction 2 moles of AlCl₃ react with 3 moles Ba(OH)₂ of molar ratio between AlCl₃ to Ba(OH)₂ is 2:3 therefore to react with 0.062832 moles of AlCl₃ required Ba(OH)₂ = 0.062832 X 3 / 2 = 0.094248 mole

moles of Ba(OH)₂ required = 0.094248 moles

molarity of Ba(OH)₂ solution = 0.664 M

Volume of solution in liter = no. of moles / molarity

volume of Ba(OH)₂ solution = 0.094248 / 0.664 = 0.14194 liter = 141.94 ml (1 L = 1000 ml then 0.14194 L = 0.14194 X 1000 = 141.94 ml)

Ans = 141.94 milliliter of Ba(OH)₂ required