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Chapter 6: Problem 63

A 100.0-mL aliquot of 0.200 \(M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

Short Answer

Expert verified
a. The balanced chemical equation: 2 KOH(aq) + Mg(NO3)2(aq) → Mg(OH)2(s) + 2 KNO3(aq) b. The precipitate that forms is Mg(OH)2. c. The mass of precipitate produced is 0.583 g. d. The concentrations of ions remaining in the solution: K⁺ ions: 0.100 M, NO3⁻ ions: 0.100 M, Mg²⁺ ions: 0 M.

Step by step solution

01

1. Determine the initial moles of each reactant

To find the initial moles of each reactant, we will use the formula: moles = concentration × volume For potassium hydroxide (KOH): moles of KOH = 0.200 M × 100.0 mL × \( \frac{1\:L}{1000\: mL} \) = 0.0200 mol For magnesium nitrate (Mg(NO3)2): moles of Mg(NO3)2 = 0.200 M × 100.0 mL × \( \frac{1\:L}{1000\: mL} \) = 0.0200 mol
02

2. Write the balanced chemical equation

The balanced chemical equation for the reaction between potassium hydroxide and magnesium nitrate: 2 KOH(aq) + Mg(NO3)2(aq) → Mg(OH)2(s) + 2 KNO3(aq)
03

3. Calculate the limiting reactant and the moles of precipitate formed

Both reactants have the same initial moles (0.0200 mol) and a stoichiometric ratio of 2:1 in the balanced equation. Therefore, the limiting reactant is potassium hydroxide (KOH). From the balanced equation, we know that 2 moles of KOH reacts completely with 1 mole of Mg(NO3)2 to form 1 mole of Mg(OH)2 precipitate. So, we can calculate the moles of precipitate formed as follows: moles of Mg(OH)2 = 0.0200 mol KOH × \( \frac{1\:mol\:Mg(OH)2}{2\:mol\:KOH} \) = 0.0100 mol Mg(OH)2
04

4. Find the mass of precipitate

To find the mass of the Mg(OH)2 precipitate, we will use the formula: mass = moles × molar mass The molar mass of Mg(OH)2 is: 24.31 (Mg) + 2 × (16.00 (O) + 1.01 (H)) = 58.33 g/mol Mass of Mg(OH)2 precipitate = 0.0100 mol × 58.33 g/mol = 0.583 g
05

5. Calculate the concentration of each ion remaining in solution

After precipitation, some of the initial ions will remain in the solution. We can calculate the moles of K⁺ ions and NO3⁻ ions remaining in solution by considering the stoichiometry of the reaction: moles of KNO3 formed = 0.0200 mol KOH × \( \frac{2\:mol\:KNO3}{2\:mol\:KOH} \) = 0.0200 mol This means 0.0200 mol of K⁺ ions and 0.0200 mol of NO3⁻ ions are in the solution. The total volume of the final solution is 200.0 mL. We can now calculate the concentrations of each ion: Concentration of K⁺ ions: \( \frac{0.0200\:mol}{0.200\:L} \) = 0.100 M Concentration of NO3⁻ ions: \( \frac{0.0200\:mol}{0.200\:L} \) = 0.100 M For the Mg²⁺ ions, all of them have reacted and formed the Mg(OH)2 precipitate, so its concentration in the solution is 0 M. Answer: a. 2 KOH(aq) + Mg(NO3)2(aq) → Mg(OH)2(s) + 2 KNO3(aq) b. Mg(OH)2 is the precipitate that forms. c. The mass of precipitate produced is 0.583 g. d. The concentration of ions remaining in the solution is: K⁺ ions: 0.100 M NO3⁻ ions: 0.100 M Mg²⁺ ions: 0 M

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Most popular questions from this chapter

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} .\) A \(0.456-\mathrm{g}\) sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{3} .\) The precipitate is filtered, dried, and weighed. Its mass is 0.107 g. Calculate the following. a. the mass of iron in the sample b. the mass of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample c. the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample

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