Answer to Question #205601 in Chemistry for marianna

Question #205601

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?


1
Expert's answer
2021-06-11T01:58:25-0400

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.

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