Write the net ionic equation for the following aqueous reaction. Indicate the spectator ions. CaCl2 (aq) + Na2SO4 (aq) → CaSO4 (s) + 2NaCl (aq) 3. Complete and balance each double displacement reaction. If no reaction occurs, write NO REACTION. FeCl2 (aq) + CaS (aq) → K2SO4 (aq) + CaI2 (aq) →
The Correct Answer and Explanation is:
Let’s tackle each question one at a time:
2. Net Ionic Equation:
The reaction provided is:
CaCl2(aq)+Na2SO4(aq)→CaSO4(s)+2NaCl(aq)\text{CaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow \text{CaSO}_4 (s) + 2\text{NaCl} (aq)CaCl2(aq)+Na2SO4(aq)→CaSO4(s)+2NaCl(aq)
Step 1: Write the full ionic equation
First, we break all soluble compounds into their ions. For this, remember that calcium chloride (CaCl₂) and sodium sulfate (Na₂SO₄) are soluble in water, while calcium sulfate (CaSO₄) is insoluble and precipitates out of the solution. Ca2+(aq)+2Cl−(aq)+2Na+(aq)+SO42−(aq)→CaSO4(s)+2Na+(aq)+2Cl−(aq)\text{Ca}^{2+} (aq) + 2\text{Cl}^- (aq) + 2\text{Na}^+ (aq) + \text{SO}_4^{2-} (aq) \rightarrow \text{CaSO}_4 (s) + 2\text{Na}^+ (aq) + 2\text{Cl}^- (aq)Ca2+(aq)+2Cl−(aq)+2Na+(aq)+SO42−(aq)→CaSO4(s)+2Na+(aq)+2Cl−(aq)
Step 2: Identify the spectator ions
Spectator ions are ions that do not undergo any chemical change and remain unchanged on both sides of the equation. In this case, the spectator ions are: Na+ and Cl−\text{Na}^+ \text{ and } \text{Cl}^-Na+ and Cl−
Step 3: Write the net ionic equation
Remove the spectator ions from the equation: Ca2+(aq)+SO42−(aq)→CaSO4(s)\text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \rightarrow \text{CaSO}_4 (s)Ca2+(aq)+SO42−(aq)→CaSO4(s)
This is the net ionic equation for the reaction. It shows that calcium ions and sulfate ions react to form solid calcium sulfate.
3. Complete and Balance the Reactions:
Reaction 1: FeCl₂ (aq) + CaS (aq)
Iron(II) chloride and calcium sulfide are both soluble, so they dissociate into ions: Fe2+(aq)+2Cl−(aq)+Ca2+(aq)+S2−(aq)→CaS(s)+FeCl2(aq)\text{Fe}^{2+} (aq) + 2\text{Cl}^- (aq) + \text{Ca}^{2+} (aq) + \text{S}^{2-} (aq) \rightarrow \text{CaS} (s) + \text{FeCl}_2 (aq)Fe2+(aq)+2Cl−(aq)+Ca2+(aq)+S2−(aq)→CaS(s)+FeCl2(aq)
This is a precipitation reaction, and CaS will form a solid since calcium sulfide is insoluble in water. The balanced equation for this reaction is: Fe2+(aq)+S2−(aq)→FeS(s)\text{Fe}^{2+} (aq) + \text{S}^{2-} (aq) \rightarrow \text{FeS} (s)Fe2+(aq)+S2−(aq)→FeS(s)
So the final equation is: FeCl2(aq)+CaS(aq)→FeS(s)+CaCl2(aq)\text{FeCl}_2 (aq) + \text{CaS} (aq) \rightarrow \text{FeS} (s) + \text{CaCl}_2 (aq)FeCl2(aq)+CaS(aq)→FeS(s)+CaCl2(aq)
Reaction 2: K₂SO₄ (aq) + CaI₂ (aq)
In this reaction, potassium sulfate (K₂SO₄) and calcium iodide (CaI₂) are both soluble. We need to check if any insoluble salts form.
Potassium and calcium ions react with sulfate and iodide ions. However, potassium iodide (KI) and calcium sulfate (CaSO₄) are generally soluble, so no precipitation occurs.
Thus, the answer is: NO REACTION\text{NO REACTION}NO REACTION
Summary of Reactions:
- FeCl₂ (aq) + CaS (aq) → FeS (s) + CaCl₂ (aq)
- K₂SO₄ (aq) + CaI₂ (aq) → NO REACTION
Explanation:
In double displacement reactions, two ionic compounds exchange their ions. If any of the resulting compounds are insoluble (do not dissolve in water), they will form a solid precipitate. In reaction 1, iron(II) and sulfide ions form an insoluble compound, iron(II) sulfide (FeS), which precipitates out. However, in reaction 2, all the products are soluble in water, so no precipitation occurs, and the reaction does not proceed.
