Calculate the mass of the precipitate obtained by reacting 160 g of sodium hydroxide (NaOH)

Calculate the mass of the precipitate obtained by reacting 160 g of sodium hydroxide (NaOH) with magnesium chloride (MgCl2).

Let’s write the reaction equation:

MgCl2 + 2NaOH = Mg (OH) 2 ↓ + 2NaCl

Find the amount of sodium hydroxide substance:

v (NaOH) = m (NaOH) / M (NaOH) = 160/40 = 4 (mol).

According to the reaction equation, 1 mol of Mg (OH) 2 is formed per 2 mol of NaOH, therefore:

v (Mg (OH) 2) = v (NaOH) / 2 = 4/2 = 2 (mol).

Thus, the mass of the obtained magnesium hydroxide precipitate is:

m (Mg (OH) 2) = v (Mg (OH) 2) * M (Mg (OH) 2) = 2 * 58 = 116 (g).

Answer: 116 g.