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 \(\mathrm{Cl}^{-}\).

To prepare hydrogen selenide (H2Se) from an aqueous acid solution and aluminum selenide (Al2Se3), the aluminum selenide reacts with an aqueous acid, generating hydrogen selenide and an aluminum salt. Here, we consider hydrochloric acid (HCl) as the aqueous acid. The equation for the reaction is: Al2Se3 + 6HCl → 2AlCl3 + 3H2Se Now we need to balance the equation: - We have 6 hydrogen atoms on the acid side, so we need 3 molecules of hydrogen selenide on the product side. - Similarly, we have 2 aluminum atoms on the reactant side, so we need 2 molecules of aluminum chloride on the product side. - Finally, we have 3 selenide atoms on the reactant side, so we need 3 molecules of hydrogen selenide on the product side. Balanced reaction: \(Al_{2}Se_{3} + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}Se\)

Sodium thiosulfate (Na2S2O3) is used to remove excess chlorine (Cl2) from bleached fabrics. The thiosulfate ion (S2O3^2-) forms sulfate ion (SO4^2-) and elemental sulfur (S), while chlorine (Cl2) is reduced to chloride ion (Cl^-). First, let's write the initial equation: Na2S2O3 + Cl2 → Na2SO4 + 2NaCl + S Now we need to balance the equation: - There are 2 sodium atoms on the reactant side, so we need 2 sodium atoms on the product side (1 in Na2SO4 and 1 in 2NaCl). - There are 2 sulfur atoms on the reactants side, so we need 1 sulfur atom in the sulfate ion and 1 sulfur atom in the elemental sulfur. - There are 3 oxygen atoms in the thiosulfate ion, so we need 4 oxygen atoms in the sulfate ion and a net increase of 1 oxygen atom. This can be achieved by including an additional water molecule (H2O) on the product side. - There are 2 chlorine atoms on the reactant side, so we need 2 chlorine atoms in the chloride ions on the product side. Balanced reaction: \(Na_{2}S_{2}O_{3} + Cl_{2} \rightarrow Na_{2}SO_{4} + 2NaCl + S + H_{2}O\)