Calculate the mass of the precipitate obtained by reacting 24 g of MgSO4 solution with excess NaOH solution.

Given:
m (MgSO4) = 24 g
To find:
m (Mg (OH) 2) =?
Decision:
1) We draw up the equation of the chemical reaction, let the mass of the sediment be x g.
MgSO4 + 2NaOH = Mg (OH) 2 + Na2SO4
2) Find the molecular weights of magnesium sulfate and magnesium hydroxide.
M (MgSO4) = 24 + 32 + 64 = 120 g / mol
M (Mg (OH) 2) = 24 + 32 + 2 = 58 g / mol
3) 24 g: 120 g / mol = x g: 58 g / mol
x = (24 x 58): 120 = 11.6 g.
Answer: m (Mg (OH) 2) = 11.6 g.