Question

Hi! Please complete and balance the following reactions, then state the molecular, complete ionic, and net ionic equations for the following reactions:

1. SnCl₄ + Pb(NO₃)₂ --->

2. NaOH + Pb(NO₃)₂ --->

3. H₂SO₄ + Ca(NO₃)₂ --->

4. HCl +  Cu(NO₃)₂  --->

5. NH₄OH + SnCl₄ --->

6. NaOH + SnCl₄ --->

7. NaOH + NH₄OH --->

8. Pb(NO₃)₂ + SnCl₄ --->

9. NaOH + Al(NO₃)₃ --->

10. NH₄OH + Al(NO₃)₃ --->

11. NaOH + Ni(NO₃)₂ --->

Thanks in advance!

Answer 1

Following is the - complete Answer -&- Explanation: for the first four sub-parts ( i.e. part - (1) to part -(4) ), of the given Question, in.....typed format....

Answer:

Following are the net ionic equations, for the given sub-parts, of the question: ( But the balanced reactions, and the complete ionic equations ) are given below, under section: "Explanation"...

1. part - (1): net ionic:   Sn4+(aq) + 4 Cl - (aq) + 2 Pb 2+ (aq) + 4 NO3 - (aq) Sn(NO3)4 (s) + 2 PbCl2 (s)
2. part - (2): net ionic:    Pb2+ (aq) + 2 OH - (aq)   Pb(OH)2 (s)
3. part - (3): net ionic:    SO42- (aq) + Ca2+ (aq)   CaSO4 (s)   
4. part - (4): net ionic:   Not Applicable in this case ...

Explanation:

Following is the complete Explanation, for the above Answer, in......typed format....

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Part - (1):

We are given with the reaction between: Tin(IV)chloride ( SnCl4 ), and Lead (II) Nitrate ( Pb(NO3)2 )

1. balanced reaction: SnCl4 (aq) + 2 Pb(NO3)2 (aq) Sn(NO3)4 (s) + 2 PbCl2 (s)
2. complete ionic: Sn4+(aq) + 4 Cl - (aq) + 2 Pb 2+ (aq) + 4 NO3 - (aq) Sn(NO3)4 (s) + 2 PbCl2 (s)

[Since, here in this case there will not be any spectator ions, following will be the net ionic equation ]

1. net ionic:   Sn4+(aq) + 4 Cl - (aq) + 2 Pb 2+ (aq) + 4 NO3 - (aq) Sn(NO3)4 (s) + 2 PbCl2 (s)

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Part - (2):

We are given with the reaction between: Sodium Hydroxide ( NaOH ), and Lead (II) Nitrate ( Pb(NO3)2 )

1. balanced reaction: Pb(NO3)2 (aq) + 2 NaOH (aq)   Pb(OH)2 (s) + 2 NaNO3 (aq)  
2. complete ionic:  Pb2+ (aq) + 2 NO3 - (aq) + 2 Na+ (aq) + 2 OH - (aq)   Pb(OH)2 (s) + 2 Na+ (aq) + 2 NO3 - (aq)

[Since, as marked above with dark brown color, the spectator ions will cancel out , on both sides, i.e. LHS and RHS, of the above equation, we will get the following net ionic equation. ]

1. net ionic:    Pb2+ (aq) + 2 OH - (aq)   Pb(OH)2 (s)

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Part - (3):

We are given with the reaction between: Sulfuric acid (H2SO4 ), and Calcium (II) Nitrate ( Ca(NO3)2 )

1. balanced reaction: H2SO4 (aq) + Ca(NO3)2 (aq)   CaSO4 (s) + 2 HNO3 (aq)
2. complete ionic:  2 H+ (aq) + SO42- (aq) + Ca2+ (aq) + 2 NO3- (aq)   CaSO4 (s) + 2 H+ (aq) + 2 NO3- (aq)  

[Since, as marked above with dark brown color, the spectator ions will cancel out , on both sides, i.e. LHS and RHS, of the above equation, we will get the following net ionic equation. ]

1. net ionic:    SO42- (aq) + Ca2+ (aq)   CaSO4 (s)   

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Part - (4):

We are given with the reaction between: Hydrochloric acid (HCl), and Copper (II) Nitrate ( Cu(NO3)2 )

1. balanced reaction: 2 HCl (aq) + Cu(NO3)2 (aq)   CuCl2 (aq) + 2 HNO3 (aq)
2. complete ionic:  2 H+ (aq) + 2 Cl - (aq) + Cu 2+ (aq) + 2 NO3- (aq)   Cu 2+ (aq) + 2 Cl - (aq) + 2 H+ (aq) + 2 NO3- (aq)

[Since, as marked above with dark brown color, the spectator ions will cancel out , on both sides, i.e. LHS and RHS, of the above equation, we will get the following net ionic equation. ]

1. net ionic:    Since, all ions will cancel out on both LHS and RHS, and all will become spectator ions, of the balanced chemical equation, there will be NO net ionic equation, available, in the given case.

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