Magnesium ions can be precipitated from seawater by the addition of sodium hydroxide.?

Mg(OH)2(s) <==> Mg2+ + 2OH- ...... Ksp = 5.61×10^-12

Mg2+ + 2NaOH(aq) --> Mg(OH)2(s) + 2Na+
0.0888g ..?g

By far, the easiest way to solve this and almost all stoichiometry problems is the unit-factor (factor label) method.

0.0888g Mg2+ x (1 mol Mg2+ / 24.3g Mg2+) x (2 mol NaOH /1 mol Mg2+) x (40.0g NaOH / 1 mol NaOH) = 0.292 g NaOH

====== Follow up ======

Chugoleung seems to have missed that the mass of magnesium in seawater is in milligrams. His answer is 1000 times too large.
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Magnesium hydroxide is soluble in water. I.e. Mg(OH)2 is insoluble and will precipitate out: Mg2+(aq) +2OH-(aq) -> Mg(OH)2 (s).

no. of mole of Mg2+ in 88.8mg of Mg2+:
mass/molar mass = 88.8/24.3 =3.65mole

According to Mg2+ +2OH- -> Mg(OH)2,
no. of mole of OH- needed = 3.65 x 2 = 7.3mole

Hence 7.3mole of OH- is needed, which is from NaOH.

How many NaOH do we need for 7.3mole of OH-?
Theres one mole of OH- in one mole of NaOH. ie, we need 7.3 mole of NaOH for 7.3mole of OH-

Mass in gram of NaOH needed = mole x molar mass = 7.3 x 40 = 292gram


hope it helps:)