Answer:
Molecular: [tex]Sr(NO_3)_2(aq) +Li_2SO_4(aq)\rightarrow SrSO_4(s)+2LiNO_3(aq)[/tex]
Ionic: [tex]Sr^{2+}(aq)+2NO_3^-(aq) +2Li^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)+2Li^+(aq)+2NO_3^-(aq)[/tex]Net ionic: [tex]Sr^{2}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)[/tex]
Explanation:
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In this case, since the molecular, ionic and net ionic equations show the complete molecules, ions and resulting ions respectively, for the reaction between strontium nitrate and lithium sulfate, we can notice the formation of solid strontium sulfate and lithium nitrate as shown below:
[tex]Sr(NO_3)_2(aq) +Li_2SO_4(aq)\rightarrow SrSO_4(s)+2LiNO_3(aq)[/tex]
Which is the molecular equation showing both reactants and products as molecules. Then, the ionic equation shows all the reactants and products as ions, considering that aqueous solutions dissociate whereas solid, liquid and gaseous molecules do not, therefore, we obtain:
[tex]Sr^{2+}(aq)+2NO_3^-(aq) +2Li^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)+2Li^+(aq)+2NO_3^-(aq)[/tex]
Finally, for the net ionic equation, we cancel out the spectator ions, which are those at both reactants and products:
[tex]Sr^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)[/tex]
Best regards!
A further explanation is below.
Given:
Strontium nitrate reacts with Lithium sulfate just to produce Strontium sulfate ([tex]Sr(NO_3)_2[/tex]) and Lithium nitrate ([tex]Li NO_3[/tex]).
The molecular equation will be:
→ [tex]Sr(NO_3)_2(aq) +Li_2SO_4(aq) \rightarrow SrSO_4 (s) +2LiNO_3 (aq)[/tex]
The complete ionic equation will be:
→ [tex]Sr^{2+} (aq) +2NO_3^- (aq) +2Li^+ (aq)+ SO_4^{2-} (aq) \rightarrow SrSO_4 (s)+2Li^+ (aq) +2NO_3^- (aq)[/tex]
By removing the uncharged ions from equation's will be:
Spectator ion:
→ [tex]2Li^+ (aq), 2NO_3^- (aq)[/tex]
Net ionic equation will be:
→ [tex]Sr^{2+}(aq)+SO_4^{2-} (aq) \rightarrow SrSO_4 (s)[/tex]
Thus the response above is right.
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