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LABORATORY UNKNOWN IDENTITY AND EQUATIONS UNKNOWN NUMBER:-5 Write molecular, ion

ID: 559726 • Letter: L

Question

LABORATORY UNKNOWN IDENTITY AND EQUATIONS UNKNOWN NUMBER:-5 Write molecular, ionic, and ner-ionic equations for the reaction of YOUR UN UNKNOWN CATION IDENTITY F ror KNOWN (it has nitrate as the anion) with cach of the cight anion test solutions. It no reaction occurs, complete the molecular and ionic equations, but write "No Reaction in place of the net-ionic equation. Underline all solids, Include the physical stare () (aq), etc.) for all species. NaCI + ionic: nct-ionic: NaBrt ionic: net-ionic: iag) Nal + (aq) ionic: net-ionic: Na,COs) 23 (a) ionic: net-ionic: Na,SOs (ag) + onic: net-ionic:

Explanation / Answer

1. Fe(NO3)3(aq) + 3 NaCl(aq) -----> FeCl3(aq) + 3 NaNO3(aq)

total ionic equation

Fe3+(aq) + 3NO3-(aq) + 3Na+(aq) + 3Cl-(aq) --------> Fe3+(aq) + 3Cl-(aq) + 3Na+(aq) + 3NO3-(aq)

Net ionic equation can be obtained by removing the ions present both side as it is:

No Reaction

Similarly for

2. 3NaBr(aq)+Fe(NO3)3(aq) -------> 3NaNO3(aq) + FeBr3(aq)

Total ionic

Fe3+(aq) + 3NO3-(aq) + 3Na+(aq) + 3Br-(aq) --------> Fe3+(aq) + 3Br-(aq) + 3Na+(aq) + 3NO3-(aq)

no net ionic reaction: FeBr3 soluble

3.

2NaI(aq) + 2 Fe(NO3)3(aq) ----------> I2(s) + 2 NaNO3(aq) + 2 Fe(NO3)2(aq)

total ionic equation

Fe3+(aq) + 3NO3-(aq) + 3Na+(aq) + 2I-(aq) --------> Fe2+(aq) + I2(s) + 2Na+(aq) + 2NO3-(aq)

net ionic equation

2I-(aq) ----->I2(s)

4. 2 Fe(NO3)3 + 3 Na2CO3 = Fe2(CO3)3 + 6 NaNO3

6Na+(aq) + 3CO32-(aq) + 2Fe+3(aq) + 6NO3-(aq) --> 6Na+(aq) + 6NO3-(aq) + Fe2(CO3)3 (s)

net ionic:
2Fe3+ (aq) + 3CO32-(aq) --> Fe2(CO3)3 (s)

5. 2 FeCl3 + 3 Na2SO4 = Fe2(SO4)3 + 6 NaCl

total ionic equation

6Na+(aq) + 3SO42-(aq) + 2Fe+3(aq) + 6NO3-(aq) --> 6Na+(aq) + 6NO3-(aq) + Fe2(SO4)3(s)

net ionic

2Fe+3(aq) + 3SO42-(aq) ------> Fe2(SO4)3(s)


2 NaI(aq) + 2 Fe(NO3)3(aq) ----------> I2(s) + 2 NaNO3(aq) + 2 Fe(NO3)2(aq)
2Na+(aq) + 2I_(aq) + 2Fe+3(aq) + 6 NO3-(aq) -------------->
2Na+(aq) + 6 NO3-(aq) + I2(s) + 2 Fe++(aq)

2Fe+3(aq) + 2I_(aq) -------> I2(s) + 2 Fe++(aq)