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The following questions may use concepts from this and previous chapters. Two beakers sit on a balance; the total mass is \(167.170 \mathrm{g} .\) One beaker contains a solution of \(\mathrm{KI}\) the other contains a solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} .\) When the solution in one beaker is poured completely into the other, the following reaction occurs: $$ 2 \mathrm{KI}(\mathrm{aq})+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{KNO}_{3}(\mathrm{aq})+\mathrm{PbI}_{2}(\mathrm{s}) $$ (IMAGE CANNOT COPY) What is the total mass of the beakers and solutions after reaction? Explain completely.

Short Answer

Expert verified
After the reaction, the total mass remains \(167.170 \mathrm{g}\).

Step by step solution

01

Understand the Reaction

The given chemical reaction is between potassium iodide (\(2 \mathrm{KI}\)) and lead nitrate (\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)). The products formed are potassium nitrate (\(2 \mathrm{KNO}_{3}\)) and lead iodide (\(\mathrm{PbI}_{2}\)). This is a typical double-displacement reaction where a solid precipitate \(\mathrm{PbI}_{2}\) is formed, but no gases are involved.
02

Apply the Law of Conservation of Mass

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction. Therefore, the total mass of reactants before the reaction should equal the total mass of products after the reaction. Since no gases are evolved or absorbed and all reactants are converted into products, the mass will remain unchanged.
03

Calculate Total Mass

The problem states that the total mass of the beakers and solutions before the reaction is \(167.170 \mathrm{g}\). According to the law of conservation of mass, this mass will remain the same after the reaction, as no mass is lost to the surroundings.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double-displacement reaction
A double-displacement reaction, also known as a metathesis reaction, involves two compounds swapping components with one another. Think of it as a classic partner exchange at a dance! The general formula for this type of reaction is: \[ AB + CD \rightarrow AD + CB \] In our exercise, the compounds are potassium iodide \((2 \mathrm{KI})\) and lead nitrate \((\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2})\). When mixed, they exchange their ions to form potassium nitrate \((2 \mathrm{KNO}_{3})\) and lead iodide \((\mathrm{PbI}_{2})\). This reaction is unique because it often results in the formation of a solid product, known as a precipitate, which in this case is \(\mathrm{PbI}_{2}\).
  • It involves exchanges between ionic compounds.
  • Usually results in the formation of a precipitate, gas, or a weak electrolyte.
This reaction is significant as it exemplifies the elegance of ionic interactions, often leading to noticeable results like precipitates in a straightforward manner.
Chemical reaction
A chemical reaction is a process where substances, known as reactants, are transformed into different substances called products. This transformation involves a change in the molecular structure, which is represented by a chemical equation. In the given exercise, the chemical equation is: \[ 2 \mathrm{KI} (\mathrm{aq}) + \mathrm{Pb}(\mathrm{NO}_{3})_{2} (\mathrm{aq}) \rightarrow 2 \mathrm{KNO}_{3} (\mathrm{aq}) + \mathrm{PbI}_{2} (\mathrm{s}) \] This equation indicates that two moles of aqueous potassium iodide react with one mole of aqueous lead nitrate to produce two moles of aqueous potassium nitrate and one mole of solid lead iodide.
  • Reactants are listed on the left side of the equation.
  • Products are listed on the right side.
  • An arrow represents the transformation from reactants to products.
Chemical reactions obey several laws, including the law of conservation of mass, which dictates that matter is neither created nor destroyed in the course of a reaction. This principle ensures that, in a closed system, the mass of the reactants will equal the mass of the products.
Precipitate formation
Precipitate formation is a hallmark characteristic of many double-displacement reactions. A precipitate is an insoluble solid that emerges from a liquid solution. In chemical terms, when two aqueous solutions react, if one of the compounds formed is insoluble in water, it separates out as a solid.
In our exercise, lead iodide \(\mathrm{PbI}_{2}\) is the precipitate. When potassium iodide and lead nitrate solutions mix, their ions rearrange, and lead iodide crystals form.
  • Identifiable visually as cloudy or solid particles in the solution.
  • Precipitation is dependent on the solubility of the product formed.
  • Often useful in various practical applications like removing contaminants from water.
The formation of a precipitate often signals a successful reaction and can help drive the reaction forward as products are removed from the solution.

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Most popular questions from this chapter

You can dissolve an aluminum soft drink can in an aqueous base such as potassium hydroxide. \(2 \mathrm{Al}(\mathrm{s})+2 \mathrm{KOH}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 2 \mathrm{KAl}(\mathrm{OH})_{4}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If you place \(2.05 \mathrm{g}\) of aluminum in a beaker with \(185 \mathrm{mL}\) of \(1.35 \mathrm{M} \mathrm{KOH},\) will any aluminum remain? What mass of \(\mathrm{KAl}(\mathrm{OH})_{4}\) is produced?

Make the following conversions. In each case, tell whether the solution is acidic or basic. \(\mathbf{p} \mathbf{H}$$\quad$$\left[\mathbf{H}_{3} \mathbf{O}^{*}\right]\) (a) 1.00\(\quad\)______ (b) 10.50\(\quad\)______ (c) ______\(\quad1.3 \times 10^{-3} \mathrm{M}\) (d) ______\(\quad2.3 \times 10^{-8} \mathrm{M}\)

Calculating and Using pH A table wine has a pH of \(3.40 .\) What is the hydronium ion concentration of the wine? Is it acidic or basic?

A An unknown metal reacts with oxygen to give the metal oxide, MO, Identify the metal if a \(0.356-\mathrm{g}\) sample of the metal produces \(0.452 \mathrm{g}\) of the metal oxide.

In some laboratory analyses, the preferred technique is to dissolve a sample in an excess of acid or base and then "back-titrate" the excess with a standard base or acid. This technique is used to assess the purity of a sample of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\). Suppose you dissolve a 0.475 -g sample of impure \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in aqueous \(\mathrm{KOH}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The NH ated in the reaction is distilled from the solution into a flask containing \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCl. The ammonia reacts with the acid to produce \(\mathrm{NH}_{4} \mathrm{Cl},\) but not all of the \(\mathrm{HCl}\) is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes \(11.1 \mathrm{mL}\) of \(0.121 \mathrm{M} \mathrm{NaOH}\). What is the weight percent of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in the \(0.475-\mathrm{g}\) sample?

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