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(1) For the precipitates that formed, write the complete molecular reaction, the

ID: 838041 • Letter: #

Question

(1) For the precipitates that formed, write the complete molecular reaction, the complete (full) ionic equation and lastly, the net ionic equation for the following:

a) BaCl2 + NaIO3 -->

b) BaCl2 + NaOH -->

c) BaCl2 + Al2(SO4)3 -->

d) Zn(NO3)2 + NaOH -->

e) NaOH + Al2(SO4)3 -->

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(2) Write a balanced molecular reaction, the complete (full) ionic equation and lastly, the net ionic equation showing the dissolution reaction for each of the precipitates that was soluble in HC2H3O2 (aq):

a) BaCl2 + NaOH -->

b) Zn(NO3)2 + NaOH -->

Please make sure to write out the molecular reactions, complete (full) equations AND the net ionic equations, thank you!

Explanation / Answer

Question 1

(a) Molecular reaction:

BaCl2 (aq) + 2NaIO3 (aq) ---> Ba(IO3)2 (s) + 2NaCl (aq)

Full ionic

Ba2+ (aq) + 2Cl- (aq) + 2Na+ (aq) + 2IO3- (Aq) ---> Ba(IO3)2 (s) + 2Cl- (aq) + 2Na+ (aq)

Net ionic

Ba2+ (aq) +  2IO3- (Aq) ---> Ba(IO3)2 (s)

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(b) Molecular:

BaCl2 (aq) + 2NaOH (aq) ---> Ba(OH)2 (s) + 2NaCl (aq)

Full ionic

Ba2+ (aq) + 2Cl- (aq) + 2Na+ (aq) + 2OH- (aq) ---> Ba(OH)2 (s) + 2Cl- (aq) + 2Na+ (aq)

Net ionic:

Ba2+ (aq) + 2OH- (aq) ---> Ba(OH)2 (s)

---

(c) Molecular:

3BaCl2 (aq) + Al2(SO4)3 (aq) ---> 3BaSO4 (s) + 2AlCl3 (aq)

Full ionic:

3Ba2+ (aq) + 6Cl- (aq) + 2Al3+ (aq) + 3SO42- (aq) --> 3BaSO4 (s) + 2Al3+ (aq) + 6Cl- (aq)

Net ionic:

3Ba2+ (aq) + 3SO42- (aq) --> 3BaSO4 (s)

---

(d) Molecular

Zn(NO3)2 (aq) + 2NaOH (aq) ---> Zn(OH)2 (s) + 2NaNO3 (aq)

Full ionic:

Zn2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + 2OH- (aq) --> Zn(OH)2 (s) + 2NO3- (aq) + 2Na+ (aq)

Net ionic:

Zn2+ (aq) + 2OH- (aq) --> Zn(OH)2 (s)

---

(e) Molecular

6NaOH (aq) + Al2(SO4)3 (aq) ---> 2Al(OH)3 (s) + 3Na2SO4 (aq)

Full ionic:

6Na+ (aq) + 6OH- (aq) + 2Al3+ (aq) + 3SO42- (aq) ---> 2Al(OH)3 (s) + 6Na+ (aq) + 3SO42- (aq)

Net ionic:

2Al3+ (aq) + 6OH- (aq)  ---> 2Al(OH)3 (s)

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Question 2

You don't really write molecular / full ionic / net ionic equations for dissolution reactions. They really just look this like :

a) The precipitate was Ba(OH)2 (s). The dissolution reaction is:

Ba(OH)2 (s) ---> Ba2+ + 2OH-

b) The precipitate was Zn(OH)2 (s). The dissolution reaction is:

Zn(OH)2 (s) ---> Zn2+ + 2OH-

Hope this was helpful!