1. Is it over yet? Mastering Chemistry Reading Exam #3 Proficiency #3
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Due Thursday by 11:59PM
Reading
Chapter 8, sections 1-4
Exam #3
Proficiency #3

2. Electrolytes Na2C2O4(aq) + CaCl2(aq)  2 NaCl(aq) + CaC2O4(s)
S-E
S-E
S-E
W-E
3 H2SO4(aq) + 2 Fe(OH)3(s)  Fe2(SO4)3(aq) + 6 H2O(l)
S-E
W-E
S-E
N-E
2 LiCl(aq) + Hg2(C2H3O2)2(aq)  2 LiC2H3O2(aq) + Hg2Cl2(s)
S-E
S-E
S-E
W-E

3. Electrolytes 2 HF(aq) + Ba(OH)2(aq)  BaF2(s) + 2 H2O(l)
W-E
S-E
W-E
N-E
K2CO3(aq) + 2 HClO4(aq)  2 KClO4(aq) + H2O(l) + CO2(g)
S-E
S-E
S-E
N-E
N-E
2 NaNO3(aq) + Mg(ClO3)2(aq)  2 NaClO3(aq) + Mg(NO3)2(aq)
S-E
S-E
S-E
S-E

4. A solution of chromium (III) sulfate is added to a solution of cesium phosphate. Give the net ionic equation for any reaction that happens.
Step 1: Write a balanced chemical equation (molecular equation)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq)
Cr2(SO4)3
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq) +
Cs3PO4 
CrPO4 +
Cs2SO4

5. Step 1: Write a balanced chemical equation (molecular equation)
A solution of chromium (III) sulfate is added to a solution of cesium phosphate. Give the net ionic equation for any reaction that happens.
Step 1: Write a balanced chemical equation (molecular equation)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq) SE SE WE SE
Step 2: Break STRONG ELECTROLYTES ONLY into ions
2 Cr+3(aq) + 3 SO42-(aq)
2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq)  2 CrPO4(s) + 6 Cs+(aq) + 3 SO42-(aq)
2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq)
2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq)  2 CrPO4(s)

6. Step 1: Write a balanced chemical equation (molecular equation)
A solution of chromium (III) sulfate is added to a solution of cesium phosphate. Give the net ionic equation for any reaction that happens.
Step 1: Write a balanced chemical equation (molecular equation)
Cr2(SO4)3(aq) + 2 Cs3PO4(aq)  2 CrPO4(s) + 3 Cs2SO4(aq) SE SE WE SE
Step 2: Break STRONG ELECTROLYTES ONLY into ions
2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq)  2 CrPO4(s) + 6 Cs+(aq) + 3 SO42-(aq)
Step 3: Cross out all spectator ions
N.I.E
2 Cr+3(aq) + 2 PO4-3(aq)  2 CrPO4(s)
2 Cr+3(aq) + 2 PO4-3(aq)  2 CrPO4(s)

7. A zinc acetate solution is combined with 3. 50 M perchloric acid
A zinc acetate solution is combined with 3.50 M perchloric acid. Give the net ionic equation.
Step 1: Write a balanced chemical equation (molecular equation)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq)
HClO4 +
Zn(C2H3O2)2 
HC2H3O2

8. Step 1: Write a balanced chemical equation (molecular equation)
A zinc acetate solution is combined with 3.50 M perchloric acid. Give the net ionic equation.
Step 1: Write a balanced chemical equation (molecular equation)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq) SE SE WE SE
Step 2: Break STRONG ELECTROLYTES ONLY into ions
2 H+(aq) + 2 ClO4-(aq)
2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq)
2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq)  2 HC2H3O2(aq) + Zn2+(aq) + 2 ClO4-(aq)
2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq)  2 HC2H3O2(aq)

9. Step 1: Write a balanced chemical equation (molecular equation)
A zinc acetate solution is combined with 3.50 M perchloric acid. Give the net ionic equation.
Step 1: Write a balanced chemical equation (molecular equation)
2 HClO4(aq) + Zn(C2H3O2)2(aq)  2 HC2H3O2(aq) + Zn(ClO4)2(aq) SE SE WE SE
Step 2: Break STRONG ELECTROLYTES ONLY into ions
2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq)  2 HC2H3O2(aq) + Zn2+(aq) + 2 ClO4-(aq)
Step 3: Cross out all spectator ions
N.I.E
2 H+(aq) + 2 C2H3O2-(aq)  2 HC2H3O2(aq)
2 H+(aq) + 2 C2H3O2-(aq)  2 HC2H3O2(aq)

10. Net Ionic Equations Start with a balanced chemical equation: M.E.
Na2C2O4(aq) + CaCl2(aq)  2 NaCl(aq) + CaC2O4(s)
Na2C2O4(aq) + CaCl2(aq)  2 NaCl(aq) + CaC2O4(s)
S.E.
S.E.
S.E.
W.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq)
2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq)
2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + CaC2O4(s)
2Na+(aq) + C2O4-2(aq)

11. C2O4-2(aq) + Ca2+(aq)  CaC2O4(s)
Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + CaC2O4(s)
Cross out every thing that is EXACTLY the same on both sides:
2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + CaC2O4(s)
N.I.E.
C2O4-2(aq) + Ca2+(aq)  CaC2O4(s)

12. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
2 HF(aq) + Ba(OH)2(aq)  BaF2(s) + 2 H2O(l)
2 HF(aq) + Ba(OH)2(aq)  BaF2(s) + 2 H2O(l)
2 HF(aq) + Ba(OH)2(aq)  BaF2(s) + 2 H2O(l)
2 HF(aq) + Ba(OH)2(aq)  BaF2(s) + 2 H2O(l)
2 HF(aq) + Ba(OH)2(aq)  BaF2(s) + 2 H2O(l)
W.E.
S.E.
W.E.
N.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 HF(aq) + Ba2+(aq) + 2OH-(aq)  BaF2(s) + 2 H2O(l)
2 HF(aq) + Ba2+(aq) + 2OH-(aq)  BaF2(s)
2 HF(aq)
2 HF(aq) + Ba2+(aq) + 2OH-(aq)

13. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 HF(aq) + Ba2+(aq) + 2OH-(aq)  BaF2(s) + 2 H2O(l)
Cross out every thing that is EXACTLY the same on both sides:
2 HF(aq) + Ba2+(aq) + 2OH-(aq)  BaF2(s) + 2 H2O(l)
N.I.E.
2 HF(aq) + Ba2+(aq) + 2OH-(aq)  BaF2(s) + 2 H2O(l)

14. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
Fe2(SO4)3(aq) + 3 PbCl2(s)  2 FeCl3(aq) + 3 PbSO4(s)
Fe2(SO4)3(aq) + 3 PbCl2(s)  2 FeCl3(aq) + 3 PbSO4(s)
Fe2(SO4)3(aq) + 3 PbCl2(s)  2 FeCl3(aq) + 3 PbSO4(s)
Fe2(SO4)3(aq) + 3 PbCl2(s)  2 FeCl3(aq) + 3 PbSO4(s)
Fe2(SO4)3(aq) + 3 PbCl2(s)  2 FeCl3(aq) + 3 PbSO4(s)
S.E.
W.E.
S.E.
W.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s)  2Fe+3(aq) + 6Cl-(aq) + 3PbSO4(s)
2Fe+3(aq) + 3SO4-2(aq)
2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s)  2Fe+3(aq) + 6Cl-(aq)
2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s)

15. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s)  2Fe+3(aq) + 6Cl-(aq) + 3PbSO4(s)
Cross out every thing that is EXACTLY the same on both sides:
2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s)  2Fe+3(aq) + 6Cl-(aq) + 3PbSO4(s)
N.I.E.
3SO4-2(aq) + 3PbCl2(s)  6Cl-(aq) + 3PbSO4(s)
3SO4-2(aq) + 3PbCl2(s)  2Cl-(aq) + 3PbSO4(s)

16. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
NaNO3(aq) + LiCl(aq)  NaCl(aq) + LiNO3(aq)
NaNO3(aq) + LiCl(aq)  NaCl(aq) + LiNO3(aq)
NaNO3(aq) + LiCl(aq)  NaCl(aq) + LiNO3(aq)
NaNO3(aq) + LiCl(aq)  NaCl(aq) + LiNO3(aq)
NaNO3(aq) + LiCl(aq)  NaCl(aq) + LiNO3(aq)
S.E.
S.E.
S.E.
S.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq) + Li+(aq) + NO3-(aq)
Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq)
Na+(aq) + NO3-(aq)
Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq)

17. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq) + Li+(aq) + NO3-(aq)
Cross out every thing that is EXACTLY the same on both sides:
Na+(aq) + NO3- (aq) + Li+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq) + Li+(aq) + NO3-(aq)
No reaction

18. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
W.E.
S.E.
S.E.
N.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq)  Ba2+(aq) + 2 C2H3O2-(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq)
2 HC2H3O2(aq)
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq)  Ba2+(aq) + 2 C2H3O2-(aq)

19. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq)  Ba2+(aq) + 2 C2H3O2-(aq) + 2 H2O(l)
Cross out every thing that is EXACTLY the same on both sides:
2 HC2H3O2(aq) + Ba2+(aq) + 2OH-(aq)  Ba2+(aq) + 2C2H3O2-(aq) + 2 H2O(l)
N.I.E.
2 HC2H3O2(aq) + 2OH-(aq)  2C2H3O2-(aq) + 2 H2O(l)
HC2H3O2(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)

20. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
S.E.
S.E.
S.E.
N.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq)  2Na+(aq) + SO4-2(aq) + 2H2O(l)
2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq)
2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq)  2Na+(aq) + SO4-2(aq)
2H+(aq) + SO4-2(aq)

21. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq)  2Na+(aq) + SO4-2(aq) + 2H2O(l)
Cross out every thing that is EXACTLY the same on both sides:
2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq)  2Na+(aq) + SO4-2(aq) + 2H2O(l)
N.I.E.
2 H+(aq) + 2 OH-(aq)  2 H2O(l)
H+(aq) + OH-(aq)  H2O(l)