/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 133 The following questions may use ... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) a base like ammonia, can react with sulfuric acid. \(2 \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2 \mathrm{N}_{2} \mathrm{H}_{5}+(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) What mass of hydrazine reacts with \(250 .\) mL. of \(0.146 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

An unknown solid acid is either citric acid or tartaric acid. To determine which acid you have, you titrate a sample of the solid with aqueous \(\mathrm{NaOH}\) and from this determine the molar mass of the unknown acid. The appropriate equations are as follows: Citric acid: \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})+3 \mathrm{NaOH}(\mathrm{aq}) \rightarrow\) $$ 3 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq}) $$ Tartanic acid: \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Na}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}(\mathrm{aq}) $$ A \(0.956-\mathrm{g}\) sample requires \(29.1 \mathrm{mL}\) of \(0.513 \mathrm{M} \mathrm{NaOH}\) to consume the acid completely. What is the unknown acid?

A pesticide contains thallium(I) sulfate, \(\mathrm{TI}_{2} \mathrm{SO}_{4}\). Dissolving a \(10.20-\mathrm{g}\) sample of impure pesticide in water and adding sodium iodide precipitates \(0.1964 \mathrm{g}\) of thallium(I) iodide, TII. $$ \mathrm{TI}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \rightarrow 2 \mathrm{TII}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) $$ What is the mass percent of \(\mathrm{TI}_{2} \mathrm{SO}_{4}\) in the original \(10.20-\mathrm{g}\) sample?

You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a \(0.251-\mathrm{g}\) sample of the alloy in acid, an excess of KI is added, and the \(\mathrm{Cu}^{2+}\) and \(1^{-}\) ions undergo the reaction $$ 2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq}) $$ The liberated \(I_{3}^{-}\) is titrated with sodium thiosulfate according to the equation \(\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})\) (a) Designate the oxidizing and reducing agents in the two reactions above. (b) If 26.32 mL of \(0.101 M N a_{2} S_{2} O_{3}\) is required for titration to the equivalence point, what is the weight percent of Cu in the alloy?

A Chromium(III) chloride forms many compounds with ammonia. To find the formula of one of these compounds, you titrate the \(\mathrm{NH}_{3}\) in the compound with standardized acid. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}} \mathrm{Cl}_{3}(\mathrm{aq})+\times \mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ x \mathrm{NH}_{4}+(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq})+(x+3) \mathrm{Cl}^{-}(\mathrm{aq}) $$ Assume that \(24.26 \mathrm{mL}\) of \(1.500 \mathrm{M} \mathrm{HCl}\) is used to tirate \(1.580 \mathrm{g}\) of \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{3} .\) What is the value of \(x ?\)

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