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What mass of \(\mathrm{KCl}\) is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution?

Short Answer

Expert verified
The mass of KCl required to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution is approximately \(0.224 \mathrm{g}\).

Step by step solution

01

List the given information and identify the target variable

We have the following information: - Volume of AgNO₃ solution = 15.0 mL - Concentration of AgNO₃ solution = 0.200 M - KCl needed to precipitate the silver ions Our target variable is the mass of KCl.
02

Convert volume to liters

Convert the volume of the AgNO₃ solution from mL to L: \(15.0 \mathrm{~mL} = 15.0 \times 10^{-3} \mathrm{~L}\)
03

Calculate moles of AgNO₃

Using the given concentration, calculate the moles of AgNO₃: Moles of AgNO₃ = Concentration × Volume = (0.200 M) × (15.0 × 10⁻³ L) Moles of AgNO₃ = 3.0 × 10⁻³ mol
04

Determine stoichiometric ratio

According to the balanced chemical equation, the stoichiometric ratio between AgNO₃ and KCl is 1:1.
05

Calculate moles of KCl

Since the stoichiometric ratio is 1:1, the moles of KCl required will be equal to the moles of AgNO₃. Moles of KCl = Moles of AgNO₃ = 3.0 × 10⁻³ mol
06

Calculate mass of KCl

Using the molar mass of KCl (39.10 g/mol for K and 35.45 g/mol for Cl), find the mass of KCl required: Mass of KCl = Moles of KCl × Molar mass of KCl Mass of KCl = (3.0 × 10⁻³ mol) × (39.10 g/mol + 35.45 g/mol) Mass of KCl = (3.0 × 10⁻³ mol) × (74.55 g/mol) Mass of KCl ≈ 0.224 g The mass of KCl required to precipitate the silver ions from 15.0 mL of 0.200 M AgNO₃ solution is approximately 0.224 g.

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Most popular questions from this chapter

\(\mathrm{HCl}, \mathrm{HBr}\), and \(\mathrm{HI}\) are strong acids, yet \(\mathrm{HF}\) is a weak acid. What does this mean in terms of the extent to which these substances are ionized in solution?

(a) Use the following reactions to prepare an activity series for the halogens: $$ \begin{array}{r} \mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\ \mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q) \end{array} $$ (b) Relate the positions of the halogens in the periodic table with their locations in this activity series. (c) Predict whether a reaction occurs when the following reagents are mixed: \(\mathrm{Cl}_{2}(a q)\) and \(\mathrm{KI}(a q) ; \mathrm{Br}_{2}(a q)\) and \(\mathrm{LiCl}(a q)\).

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M}\) HCl. The \(\mathrm{NH}_{3}\) reacts with HCl as follows: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min}\), the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) ls this manufacturer in compliance with regulations?

Write the balanced molecular and net ionic equations for each of the following neutralization reactions: (a) Aqueous acetic acid is neutralized by aqueous potassium hydroxide. (b) Solid chromium(III) hydroxide reacts with nitric acid. (c) Aqueous hypochlorous acid and aqueous calcium hydroxide react.

A sample of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) is stirred in water at \(30^{\circ} \mathrm{C}\) until the solution contains as much dissolved \(\mathrm{Ca}(\mathrm{OH})_{2}\) as it can hold. A \(100-\mathrm{mL}\) sample of this solution is withdrawn and titrated with \(5.00 \times 10^{-2} \mathrm{M} \mathrm{HBr}\). It requires \(48.8 \mathrm{~mL}\) of the acid solution for neutralization. What is the molarity of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution? What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, at \(30^{\circ} \mathrm{C}\), in grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) per \(100 \mathrm{~mL}\) of solution?

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