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Chapter 20: Problem 44

Write equations describing the reactions of Sn with each of the following: \(\mathrm{Cl}_{2}, \mathrm{O}_{2},\) and \(\mathrm{HCl}\) .

Short Answer

Expert verified
The balanced chemical equations for the reactions between tin (Sn) and the given reactants are: 1. Reaction with chlorine gas (Cl鈧�): \[Sn_{(s)} + 2Cl_{2(g)} \rightarrow SnCl_{4(s)}\] 2. Reaction with oxygen gas (O鈧�): \[2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{2(s)}\] 3. Reaction with hydrochloric acid (HCl): \[Sn_{(s)} + 2HCl_{(aq)} \rightarrow SnCl_{2(aq)} + H_{2(g)}\]

Step by step solution

01

Reaction with Chlorine gas (Cl鈧�)

Tin can react with chlorine gas to form tin(IV) chloride (SnCl鈧�). The balanced equation for this reaction can be written as: \[Sn_{(s)} + 2Cl_{2(g)} \rightarrow SnCl_{4(s)}\]
02

Reaction with Oxygen gas (O鈧�)

Tin can react with oxygen gas to form tin(IV) oxide (SnO鈧�). The balanced equation for this reaction can be written as: \[2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{2(s)}\]
03

Reaction with Hydrochloric Acid (HCl)

Tin can react with hydrochloric acid to form tin(II) chloride (SnCl鈧�) and hydrogen gas (H鈧�). The balanced equation for this reaction can be written as: \[Sn_{(s)} + 2HCl_{(aq)} \rightarrow SnCl_{2(aq)} + H_{2(g)}\]

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Key Concepts

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

Tin Reactions
Tin, represented by the chemical symbol Sn, is a versatile metal known for forming a variety of compounds through chemical reactions. When tin reacts with different substances such as chlorine, oxygen, and hydrochloric acid, unique and important chemical transformations take place.
  • **Tin and Chlorine Reaction:** When tin comes in contact with chlorine gas (\(Cl_2\)), it forms tin(IV) chloride, \(SnCl_4\), a compound often used in creating advanced materials.
  • **Tin and Oxygen Reaction:** Tin also reacts with oxygen gas (\(O_2\)) to produce tin(IV) oxide, \(SnO_2\). This white, powdery solid serves in various applications like in sensors and ceramics.
  • **Tin and Hydrochloric Acid Reaction:** When tin is introduced to hydrochloric acid (\(HCl\)), a reaction occurs yielding tin(II) chloride, \(SnCl_2\), alongside hydrogen gas. This reaction is notable for its role in producing important tin salts used in dyeing and printing.
Understanding tin reactions not only helps in grasping fundamental inorganic chemistry principles but also sheds light on their practical applications in industry.
Balanced Chemical Equations
Balanced chemical equations are crucial in chemistry as they reflect the conservation of mass in chemical reactions. The number of each type of atom on the reactant side must equal the number on the product side.
  • For instance, the reaction of tin with chlorine is expressed as: \[Sn_{(s)} + 2Cl_{2(g)} \rightarrow SnCl_{4(s)}\]
  • Similarly, the interaction between tin and oxygen gas is shown in the balanced equation: \[2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{2(s)}\]
  • Lastly, the combination of tin and hydrochloric acid can be represented by: \[Sn_{(s)} + 2HCl_{(aq)} \rightarrow SnCl_{2(aq)} + H_{2(g)}\]
By ensuring chemical equations are balanced, chemists can predict the amounts of products formed and the reactants needed in a given reaction. This balance is foundational for conducting experiments and industrial processes safely and efficiently.
Inorganic Chemistry
Inorganic chemistry, a branch of chemistry distinct from organic chemistry, deals with the properties and behavior of inorganic compounds, including metals, minerals, and organometallics. Tin, being a metal, comes under the purview of inorganic chemistry.
  • **Metal Reactions:** Tin exhibits several interesting reactions with non-metals and acids, as seen with its interactions with \(Cl_2, \)\(O_2, \) and \(HCl\).
  • **Compound Formation:** By understanding these reactions, students gain insights into compound formation, redox reactions, and the reactivity of different elements.
  • **Industrial Application:** Inorganic compounds formed through such reactions are vital in numerous industries, from electronics to textiles.
Studying inorganic chemistry provides the tools to understand and manipulate substances that do not revolve around carbon-based compounds, opening a vast field of scientific and practical innovation.

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Most popular questions from this chapter

While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in Te(OH) \(_{6} ?\) b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$ \mathrm{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q) $$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas: $$ \mathrm{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \mathrm{TeF}_{6}(g) $$ If a cubic block of tellurium (density \(=6.240 \mathrm{g} / \mathrm{cm}^{3} )\) measuring 0.545 \(\mathrm{cm}\) on edge is allowed to react with 2.34 \(\mathrm{L}\) fluorine gas at 1.06 \(\mathrm{atm}\) and \(25^{\circ} \mathrm{C},\) what is the pH of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated TeF \(_{6}(g)\) in 115 \(\mathrm{mL}\) solution? Assume 100\(\%\) yield in all reactions.

Calculate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\left(K_{\mathrm{sp}}=8.9 \times 10^{-12}\right)\) in an aqueous solution buffered at \(\mathrm{pH}=9.42 .\)

The electrolysis of aqueous sodium chloride (brine) is an important industrial process for the production of chlorine and sodium hydroxide. In fact, this process is the second largest consumer of electricity in the United States, after the production of aluminum. Write a balanced equation for the electrolysis of aqueous sodium chloride (hydrogen gas is also produced).

The atmosphere contains \(9.0 \times 10^{-6 \%} \mathrm{Xe}\) by volume at 1.0 atm and \(25^{\circ} \mathrm{C} .\) a. Calculate the mass of \(\mathrm{Xe}\) in a room 7.26 \(\mathrm{m}\) by 8.80 \(\mathrm{m}\) by 5.67 \(\mathrm{m} .\) b. A typical person takes in about 2 \(\mathrm{L}\) of air during a breath. How many Xe atoms are inhaled in each breath?

Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triiodide (NI \(_{3} )\) is known to be a highly explosive material. \(\mathrm{NI}_{3}\) can be synthesized according to the equation $$ \mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g) $$ a. What is the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) given the enthalpy of reaction \((-307 \mathrm{kJ})\) and the enthalpies of formation for \(\mathrm{BN}(s)(-254 \mathrm{kJ} / \mathrm{mol}), \mathrm{IF}(g)(-96 \mathrm{kJ} / \mathrm{mol})\) and \(\mathrm{BF}_{3}(g)(-1136 \mathrm{kJ} / \mathrm{mol}) ?\) b. It is reported that when the synthesis of \(\mathrm{NI}_{3}\) is conducted using 4 moles of IF for every 1 mole of \(\mathrm{BN}\) , one of the by-products isolated is \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .\) What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?
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