Answer:
Concentration of NH₄⁺ = 6.00 mol/L
Concentration of S²⁻ = 3.00 mol/L
(NH₄)₂S is an ionic compound, with the name Ammonium Sulfide. Ionic compounds are composed of ions held together by electrostatic forces, known as ionic bonding.
The compound is neutral overall, but consists of positively charged ions called cations, and negatively charged ions called anions.
Ions are atoms or molecules that have lost or gained electrons, resulting in a net charge. The cations are attracted to the anions because opposite charges attract.
Ammonium Sulfide is composed of the ammonium cation (NH₄⁺), and the sulfide anion (S²⁻). The following ionic equation represents the dissolution reaction of solid ammonium sulfide into its individual ions upon reaction with water.
[tex]\boxed{\Large \textsf{$\rm (NH_4)_2S_{\,(s)} \leftrightharpoons 2NH_4^{\ \ +}{}_{(aq)}+S^{2-}_{\ \ \ \ \,(aq)}$}}[/tex]
Since we are not given the volume of the solution, we can provide an arbitrary volume, as the volume will remain constant in the solution. Let us assume the volume of solution is 1 litre.
By this assumption, the reagents will remain in stoichiometric ratios (molar ratio of reactants to products), and therefore 1 mole of (NH₄)₂S will dissolve to produce 2 moles of NH₄⁺ and 1 mole of S²⁻.
Since we have 3.00 moles per litre, and we have 1 litre, therefore there are 3.00 moles of (NH₄)₂S in solution.
By stoichiometry:
Moles of NH₄⁺ = 3.00 × 2 = 6.00 mol
Moles of S²⁻ = 3.00 mol
Therefore, in 1 litre:
Concentration of NH₄⁺ = 6.00 mol/L
Concentration of S²⁻ = 3.00 mol/L
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