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Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

Short Answer

Expert verified
The balanced chemical equations are: (a) FeS(s) + 2HCl(aq) → H2S(g) + FeCl2(aq), (b) 4HCl(aq) + MnO2(s) → Cl2(g) + MnCl2(aq) + 2H2O(l), (c) 8NH3 + 3Br2 → N2 + 6NH4Br, and (d) Ba(ClO2)2 + H2SO4 → 2HClO2 + BaSO4(s).

Step by step solution

01

Preparation of H2S(g)

HCl(aq) and FeS(s) react to form H2S(g) and a by-product, FeCl2(s). As per stoichiometry, one molecule of HCl reacts with one molecule of FeS to produce one molecule of H2S and one molecule of FeCl2. Therefore, the balanced chemical equation is: \\[FeS(s) + 2HCl(aq) \longrightarrow H2S(g) + FeCl2(aq)\]
02

Preparation of Cl2(g)

When HCl(aq) is heated with MnO2(s), it leads to the formation of Cl2(g), MnCl2(aq) and H2O(l). Balancing the equation yields: \\[4HCl(aq) + MnO2(s) \longrightarrow Cl2(g) + MnCl2(aq) + 2H2O(l)\]
03

Preparation of N2

NH3 and Br2 react to form N2 and NH4Br. Balancing the equation gives: \\[8NH3 + 3Br2 \longrightarrow N2 + 6NH4Br\]
04

Preparation of chlorous acid

When an aqueous suspension of barium chlorite is treated with dilute H2SO4(aq), chlorous acid and barium sulfate are formed. The balanced equation for this reaction is: \\[Ba(ClO2)2 + H2SO4 \rightarrow 2HClO2 + BaSO4(s)\]

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Key Concepts

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

Balanced Chemical Equations
In chemistry, balanced chemical equations are like detailed recipes that tell us how substances react to form products. They are essential for understanding chemical reactions because they show the exact proportions of reactants and products involved. To be balanced, the number of atoms for each element must be the same on both sides of the equation. This balancing is achieved through stoichiometry, which ensures the conservation of mass.

For example, consider the reaction between hydrochloric acid (HCl) and iron sulfide (FeS) to produce hydrogen sulfide (H2S) and iron(II) chloride (FeCl2). In the balanced equation:
  • The reaction is: \[FeS(s) + 2HCl(aq) \rightarrow H2S(g) + FeCl2(aq)\]
  • Here, each side of the equation has 1 Fe, 1 S, and 2 Cl atoms.
  • Balancing the equation ensures that the chemical reaction complies with the law of conservation of mass.
By understanding balanced chemical equations, chemists can predict how a reaction will proceed and what amounts of substance are necessary to react completely.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the relationship between the quantities of reactants and products in chemical reactions. It helps us determine the proportions in which chemicals combine and how much product will form. By using a balanced chemical equation, we can perform calculations to predict these quantities accurately.

Key principles in stoichiometry include:
  • Moles: The mole is a basic unit in chemistry that represents a specific number of molecules or atoms. Stoichiometry relies heavily on mole calculations.
  • Proportions: Using the coefficients in a balanced chemical equation, stoichiometry ensures that we maintain the correct proportions of reactants and products.
  • Conversions: Stoichiometry involves converting masses or volumes using molar masses or gas laws to find out how much of a substance is consumed or produced.
Understanding stoichiometry ensures that reactions are both efficient and predictable, allowing chemists to create desired amounts of products safely.
Laboratory Methods for Chemical Preparation
Laboratory methods for chemical preparation involve specific techniques and reactions to synthesize small quantities of chemicals in a lab setting. This involves precise measurements and controlled conditions to ensure the successful reproduction of a reaction on a small scale.

Here are a few methods from the exercise:
  • Synthesis of Gases: Reactions such as \(FeS + 2HCl \rightarrow H2S + FeCl2\) show how gases like hydrogen sulfide can be produced from solid and liquid reactants.
  • Reactions with Heat: Some preparations, such as producing chlorine gas \(4HCl + MnO2 \rightarrow Cl2 + MnCl2 + 2H2O\), require heat to initiate or speed up the reaction.
  • Aqueous Reactions: These reactions happen in a solution. For instance, the preparation of nitrogen involves reacting ammonia with bromine in water \(8NH3 + 3Br2 \rightarrow N2 + 6NH4Br\).
These methods are fundamental for chemists to understand chemical behavior in controlled environments, guiding not only safe laboratory practices but also industrial chemical production.

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Most popular questions from this chapter

Complete and balance these half-equations. (a) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{CO}_{2}\) (acidic solution) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}\) (acidic solution) (c) \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{MnO}_{2}\) (basic solution) Indicate whether oxidation or reduction is involved.

The active component in one type of calcium dietary supplement is calcium carbonate. A \(1.2450 \mathrm{g}\) tablet of the supplement is added to \(50.00 \mathrm{mL}\) of \(0.5000 \mathrm{M} \mathrm{HCl}\) and allowed to react. After completion of the reaction, the excess HCl(aq) requires \(40.20 \mathrm{mL}\) of \(0.2184 \mathrm{M}\) NaOH for its titration to the equivalence point. What is the calcium content of the tablet, expressed in milligrams of \(\mathrm{Ca}^{2+} ?\)

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) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

An unknown white solid consists of two compounds, each containing a different cation. As suggested in the illustration, the unknown is partially soluble in water. The solution is treated with \(\mathrm{NaOH}(\mathrm{aq})\) and yields a white precipitate. The part of the original solid that is insoluble in water dissolves in \(\mathrm{HCl}(\mathrm{aq})\) with the evolution of a gas. The resulting solution is then treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and yields a white precipitate. (a) Is it possible that any of the cations \(M g^{2+}, C u^{2+}\) \(\mathrm{Ba}^{2+}, \mathrm{Na}^{+},\) or \(\mathrm{NH}_{4}^{+}\) were present in the original unknown? Explain your reasoning. (b) What compounds could be in the unknown mixture (that is, what anions might be present)?

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) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\).

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