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When treated with dilute \(\mathrm{HCl}(\mathrm{aq}),\) the solid that reacts to produce a gas is (a) \(\mathrm{BaSO}_{3} ;\) (b) \(\mathrm{ZnO};\) (c) \(\mathrm{NaBr} ;\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\).

Short Answer

Expert verified
The solid that reacts with dilute hydrochloric acid (HCl) to produce a gas is (a) Barium Sulphite (BaSO3).

Step by step solution

01

Understanding Reactions

Understand the reactions of different salts with dilute hydrochloric acid. Barium Sulphite (BaSO3) reacts with dilute HCl to produce Barium Chloride, Sulphur Dioxide and water. The reaction can be written as: \[ BaSO_{3(s)} + 2HCl_{(aq)} \to BaCl_{2(aq)} + SO_{2(g)} + H_{2O(l)}\]. \nZinc Oxide (ZnO) dissolves in HCl to form Zinc Chloride and Water: \[ ZnO_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_{2O(l)} \] \nSodium Bromide (NaBr) reacts with HCl and no change occurs because NaBr and HCl are both salts of strong acids and bases: \[ NaBr_{(s)} + HCl_{(aq)} \to NaCl_{(aq)} + HBr_{(aq)} \] \nSodium Sulphate (Na2SO4) also reacts with HCl, but no gas producing reaction occurs: \[ Na_{2}SO_{4(s)} + 2HCl_{(aq)} \to 2NaCl_{(aq)} + H_{2}SO_{4(aq)} \]
02

Identifying the Solid That Produces Gas

From the step 1 reactions, it's clear that the reaction producing a gas is the one involving Barium Sulphite (BaSO3).

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Key Concepts

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

Barium Sulphite Reaction
When barium sulphite ((BaSO_{3})) comes into contact with hydrochloric acid (HCl), a chemical reaction occurs where the barium sulphite dissolves, and new substances are formed. The products of this reaction are barium chloride ((BaCl_{2})), sulphur dioxide gas ((SO_{2})), and liquid water ((H_{2}O)). Let's break down what happens.
During the reaction, the acid provides hydrogen ions ((H^+)) that interact with the barium sulphite, leading to its dissolution. This process breaks the strong bonds in (BaSO_3), releasing sulphur dioxide as a gas — this is a telltale sign of a reaction occurring. It's important to remember that sulphur dioxide is a pungent, colorless gas and should be handled with proper ventilation due to its toxic nature.

Visual Clues of the Reaction

  • Bubbling or fizzing indicating gas evolution
  • Change in the color of the solution if impurities are present
  • Formation of a clear solution indicating the solubility of products
Understanding the Barium Sulphite reaction is crucial as it highlights the concept of a double displacement reaction, where ions switch partners to form new compounds.
Gas Evolution Reaction
A gas evolution reaction is a type of chemical reaction during which one of the products is a gas. In our example with barium sulphite and hydrochloric acid, sulphur dioxide ((SO_{2})) is the gas produced. These reactions are generally quite visible, as they often produce bubbles or effervescence and sometimes even a distinct smell, as with the pungent odor of sulphur dioxide.

Characteristics of Gas Evolution Reactions

  • Release of a gas, creating bubbles
  • Change in the smell due to the new gas evolved
  • Change in pH if the gas interacts with water (acidic or basic gases)
Gas evolution reactions are essential in both academic and industrial settings, as they can be indicative of reaction progress or used in processes like fermentation, where carbon dioxide is produced, or in environmental scrubbing technologies that remove toxic gases.
Solubility of Salts in Acid
Solubility refers to how well a substance (the solute) can dissolve in a solvent to form a homogeneous mixture. When it comes to salts in acids, solubility greatly depends on the nature of the salt and the acid. Strong acids, such as hydrochloric acid, can dissolve a wide range of salts by breaking the ionic bonds in solid salts, allowing the ions to disperse throughout the solution.
Solubility is a key concept in chemical reactions because it determines whether a reaction will occur if the reactants are actually able to interact in solution. For instance, barium sulphite ((BaSO_{3})) is soluble in the presence of hydrochloric acid. However, not all salts react with acids to form soluble compounds — some may form a precipitate, which is an insoluble solid that settles out of the solution.

Factors Influencing Solubility in Acid

  • The temperature of the solvent
  • The concentration of the acid
  • The presence of other ions in solution that can form complex ions
An understanding of solubility is essential in predicting the direction of a reaction and the state of the products formed.

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Most popular questions from this chapter

We want to determine the acetylsalicyclic acid content of a series of aspirin tablets by titration with \(\mathrm{NaOH}(\mathrm{aq})\) Each of the tablets is expected to contain about \(0.32\) \(\mathrm{g}\) of \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4} \cdot\) What molarity of \(\mathrm{NaOH}(\mathrm{aq})\) should we use for titration volumes of about \(23\) \(\mathrm{mL}\) ? (This procedure ensures good precision and allows the titration of two samples with the contents of a 50 mL buret.) \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow_{\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{O}_{4}^{-}}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

Assuming the volumes are additive, what is the \(\left[\mathrm{NO}_{3}^{-}\right]\) in a solution obtained by mixing \(275 \mathrm{mL}\) of \(0.283 \mathrm{M} \mathrm{KNO}_{3}, 328 \mathrm{mL}\) of \(0.421 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2},\) and \(784 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)

A \(\mathrm{KMnO}_{4}(\) aq) solution is to be standardized by titration against \(\mathrm{As}_{2} \mathrm{O}_{3}(\mathrm{s}) .\) A \(0.1078 \mathrm{g}\) sample of \(\mathrm{As}_{2} \mathrm{O}_{3}\) requires \(22.15 \mathrm{mL}\) of the \(\mathrm{KMnO}_{4}(\) aq) for its titration. What is the molarity of the \(\mathrm{KMnO}_{4}(\) aq)? \(5 \mathrm{As}_{2} \mathrm{O}_{3}+4 \mathrm{MnO}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O}+12 \mathrm{H}^{+} \longrightarrow\) \(10 \mathrm{H}_{3} \mathrm{AsO}_{4}+4 \mathrm{Mn}^{2+}\)

The following reactions do not occur in aqueous solutions. Balance their equations by the half-equation method, as suggested in Are You Wondering \(5-2\) (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \longrightarrow\) \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

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