Predict the products of the following solutions. Write the molecular, ionic, and net ionic equations for the reactions: Cr(NO3)3 and Na2SO4 Molecular Equation: Cr(NO3)3 + 3Na2SO4 -> Cr2(SO4)3 + 6NaNO3 Ionic Equation: 2Cr3+ + 6NO3- + 6Na+ + 3SO4^2- -> Cr2(SO4)3 + 6Na+ + 6NO3- Net Ionic Equation: 2Cr3+ + 3SO4^2- -> Cr2(SO4)3
The Correct Answer and Explanation is:
To predict the products of the reaction between Cr(NO3)3 and Na2SO4, let’s go step by step.
1. Writing the Molecular Equation
The molecular equation shows the full chemical formula of all reactants and products involved in the reaction, with no emphasis on the dissociation of ions in solution.
The reaction involves chromium(III) nitrate (Cr(NO3)3) and sodium sulfate (Na2SO4), both of which are ionic compounds. When mixed in aqueous solutions, they undergo double displacement (precipitation) reactions. Here’s how we can write the molecular equation:
- Chromium(III) nitrate dissociates as Cr3+ and NO3- ions in solution.
- Sodium sulfate dissociates as Na+ and SO4^2- ions in solution.
In this reaction, chromium(III) ions (Cr3+) will pair with sulfate ions (SO4^2-) to form chromium(III) sulfate (Cr2(SO4)3), which is insoluble and precipitates out of the solution. Meanwhile, sodium ions (Na+) will pair with nitrate ions (NO3-) to form sodium nitrate (NaNO3), which remains soluble in solution.
So, the molecular equation is:Cr(NO3)3(aq)+3Na2SO4(aq)→Cr2(SO4)3(s)+6NaNO3(aq)\text{Cr(NO}_3\text{)}_3 (aq) + 3\text{Na}_2\text{SO}_4 (aq) \rightarrow \text{Cr}_2\text{(SO}_4\text{)}_3 (s) + 6\text{NaNO}_3 (aq)Cr(NO3)3(aq)+3Na2SO4(aq)→Cr2(SO4)3(s)+6NaNO3(aq)
2. Writing the Ionic Equation
The ionic equation breaks down the soluble compounds into their constituent ions. It helps to see which ions participate in the reaction.
- Cr(NO3)3 dissociates into 2Cr3+2 \text{Cr}^{3+}2Cr3+ and 6NO3−6 \text{NO}_3^-6NO3−.
- Na2SO4 dissociates into 6Na+6 \text{Na}^+6Na+ and 3SO42−3 \text{SO}_4^{2-}3SO42−.
Thus, the ionic equation is:2Cr3+(aq)+6NO3−(aq)+6Na+(aq)+3SO42−(aq)→Cr2(SO4)3(s)+6Na+(aq)+6NO3−(aq)2\text{Cr}^{3+} (aq) + 6\text{NO}_3^- (aq) + 6\text{Na}^+ (aq) + 3\text{SO}_4^{2-} (aq) \rightarrow \text{Cr}_2\text{(SO}_4\text{)}_3 (s) + 6\text{Na}^+ (aq) + 6\text{NO}_3^- (aq)2Cr3+(aq)+6NO3−(aq)+6Na+(aq)+3SO42−(aq)→Cr2(SO4)3(s)+6Na+(aq)+6NO3−(aq)
3. Writing the Net Ionic Equation
The net ionic equation removes the spectator ions (ions that do not change during the reaction) from the ionic equation. Here, the spectator ions are sodium (Na+) and nitrate (NO3-), as they remain in the aqueous phase and do not participate in the formation of the precipitate.
After removing these spectator ions, the net ionic equation is:2Cr3+(aq)+3SO42−(aq)→Cr2(SO4)3(s)2\text{Cr}^{3+} (aq) + 3\text{SO}_4^{2-} (aq) \rightarrow \text{Cr}_2\text{(SO}_4\text{)}_3 (s)2Cr3+(aq)+3SO42−(aq)→Cr2(SO4)3(s)
Explanation:
- Cr(NO3)3 is a soluble ionic compound that dissociates into Cr3+ and NO3- ions in aqueous solution.
- Na2SO4 also dissociates into Na+ and SO4^2- ions in aqueous solution.
- The Cr3+ ions react with SO4^2- ions to form Cr2(SO4)3, which is insoluble in water and precipitates out of the solution.
- The Na+ and NO3- ions remain in the solution as spectator ions, not participating in the reaction.
This results in the net ionic equation showing only the formation of the chromium sulfate precipitate.
