Answer to Question #205601 in Chemistry for marianna

Question

How many grams of solid will precipitate if 26.0 mL of 0.285 mol/L magnesium nitrate are combined with 32.0 mL of 0.265 mol/L potassium hydroxide?

Answer 1

Solution:

magnesium nitrate - Mg(NO₃)₂

potassium hydroxide - KOH

Balanced chemical equation:

Mg(NO₃)₂(aq) + 2KOH(aq) → Mg(OH)₂(s) + 2KNO₃(aq)

According to the equation above: n(Mg(NO₃)₂) = n(KOH)/2 = n(Mg(OH)₂)

26.0 mL of a 0.285 M solution of Mg(NO₃)₂ contains:

n(Mg(NO₃)₂) = (26.0 mL) × (1 L / 1000 mL) × (0.285 mol / 1 L) = 0.00741 mol

32.0 mL of a 0.265 M solution of KOH contains:

n(KOH) = (32.0 mL) × (1 L / 1000 mL) × (0.265 mol / 1 L) = 0.00848 mol

Determine which reactant is limiting:

if KOH is limiting it would react with (0.00848 mol / 2) = 0.00424 mol of Mg(NO₃)₂

The initial amount of Mg(NO₃)₂ (0.00741 mol) is more than this.

Therefore, KOH is limiting reactant.

Thus, based on the amount of KOH present, the maximum amount of Mg(OH)₂ formed is:

n(Mg(OH)₂) = n(KOH) / 2 = 0.00848 mol / 2 = 0.00424 mol

The molar mass of Mg(OH)₂ is 58.32 g/mol.

Therefore,

(0.00424 mol Mg(OH)₂) × (58.32 g Mg(OH)₂ / 1 mol Mg(OH)₂) = 0.247 g Mg(OH)₂

Answer: 0.247 grams of solid will precipitate.