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Chapter 22: Problem 48

Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess \(\mathrm{Cl}_{2}\) from chlorine-bleached fabrics. The thiosulfate ion forms \(\mathrm{SO}_{4}^{2-}\) and elemental sulfur, while \(\mathrm{Cl}_{2}\) is reduced to \(\mathrm{Cl}^{-}\).

Short Answer

Expert verified
The balanced chemical equations for the given reactions are: (a) 6HCl + Al2Se3 -> 3H2Se + 2AlCl3 (b) Na2S2O3 + 4Cl2 -> Na2SO4 + 8S + 2Cl^-

Step by step solution

01

(a) Identify reactants and products

For this reaction, we have: - Reactants: Aqueous acid solution (Hydrochloric acid - HCl) and Aluminum Selenide (Al2Se3) - Product: Hydrogen selenide (H2Se) Now let's write the unbalanced equation: HCl + Al2Se3 -> H2Se + ? We need to identify the second product, which consists of the remaining elements, aluminum and chlorine. As they can form a compound, we expect the second product to be aluminum chloride (AlCl3). The unbalanced equation is now: HCl + Al2Se3 -> H2Se + AlCl3
02

(a) Balance the chemical equation

To balance this equation, we need to make sure the number of atoms for each element is the same on both sides of the equation. We have 2 Aluminum atoms, 3 Selenium atoms, and 1 Chlorine atom in the reactant side, and we need to adjust the coefficients accordingly. The balanced chemical equation is: 6HCl + Al2Se3 -> 3H2Se + 2AlCl3
03

(b) Identify reactants and products

For this reaction, we have: - Reactants: Sodium thiosulfate (Na2S2O3) and Chlorine (Cl2) - Products: Sulfate ions (SO4^2-), Elemental sulfur (S), and Chloride ions (Cl-) To neutralize the charges, we make sulfate ions (SO4^2-) part of a compound with sodium ions (Na+). So, the product is sodium sulfate (Na2SO4). Now, let's write the unbalanced equation: Na2S2O3 + Cl2 -> Na2SO4 + S + Cl^-
04

(b) Balance the chemical equation

To balance this equation, we need to make sure the number of atoms for each element is the same on both sides of the equation. We have 2 Sodium atoms, 2 Sulfur atoms, 3 Oxygen atoms, and 2 Chlorine atoms in the reactant side, and we need to adjust the coefficients accordingly. The balanced chemical equation is: Na2S2O3 + 4Cl2 -> Na2SO4 + 8S + 2Cl^-

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Most popular questions from this chapter

Name the following compounds and assign oxidation states to the halogens in them: (a) \(\mathrm{BCl}_{3}\), (b) \(\mathrm{Sr}\left(\mathrm{IO}_{4}\right)_{2}\), (c) \(\mathrm{LiOCl}\) (d) \(\mathrm{HClO}\) (e) \(\mathrm{CuClO}\) (f) \(\mathrm{Mg}\left(\mathrm{IO}_{2}\right)\).

(a) Determine the number of sodium ions in the chemical formula of albite, \(\mathrm{Na}_{x} \mathrm{AlSi}_{3} \mathrm{O}_{8} .(\mathbf{b})\) Determine the number of hydroxide ions in the chemical formula of tremolite, $\mathrm{Ca}_{2} \mathrm{Mg}_{5}\left(\mathrm{Si}_{4} \mathrm{O}_{11}\right)_{2}(\mathrm{OH})_{x}$.

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

Consider the elements $\mathrm{Ba}, \mathrm{Na}, \mathrm{O}, \mathrm{B}, \mathrm{P},\( and \)\mathrm{Kr}$. From this list, select the element that (a) is most electronegative, (b) has the greatest metallic character, \((\mathbf{c})\) most readily forms a positive ion, \((\mathbf{d})\) exhibits a maximum oxidation sate of +5 , (e) exists as monoatomic gas at room temperature, (f) has multiple allotropes.

Nitric acid is a powerful oxidizing agent. Using standard reduction potentials, predict whether the following metals can be oxidized to +2 ions by nitric acid: (a) iron, (b) copper, (c) rhodium, (d) zinc, (e) lead, (f) tin.
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