PLEASE HELP CHEMISTRY LAB WORK DUE MONDAY MORNING, KINDLY SHOW ALL WORKINGS AND

Question

PLEASE HELP CHEMISTRY LAB WORK DUE MONDAY MORNING, KINDLY SHOW ALL WORKINGS AND CALCULATIONS

TOPIC: METATHESIS REACTIONS AND NET IONIC EQUATIONS.

Write the Molecular equation, Complete Ionic equation, and Net Ionic equation for the following reactions:

(1). Nickel chloride + Sodium carbonate

(2). Hydrochloric acid + Sodium hydroxide

(3). Ammonium chloride + Sodium hydroxide

(4). Sodium acetate + Hydrochloric acid

(5). Sodium sulfide + Hydrochloric acid

(6). Lead nitrate + Sodium sulfide

(7). Lead nitrate + Sulfuric acid

(8). Potassium chloride + Sodium nitrate

Explanation / Answer

2

1)The molecular equation is as follows:

HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

2. The complete ionic equation (the one you're looking for) consists of all soluble solutes completely dissolved in solution and is as follows:

H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) --> H2O (l) + Na+ (aq) + Cl- (aq)

3. The net ionic equation is the same as the total, or complete, ionic equation except that the spectator ions are left out. Spectator ions are those that appear on both the product side and the reactant side.

H+ (aq) + OH- (aq) --> H2O (l)

3. molecular equation

NH4Cl(aq) + NaOH(aq) -----> NH3(g) + NaCl(aq) + H2O(l)

Every molecule above with an (aq) next to it is in aqueous solution, and exists in the solution as its constituent ions. The full ionic equation shows all ionic species in the reaction

Full ionic equation

NH4(+)(aq) + Cl(-)(aq) + Na(+)(aq) + OH(-)(aq) -----> NH3(g) + Na(+)(aq) + Cl(-)(aq) + H2O(l)

Now, you can see that some of the ionic species are present on both sides of the equation (Na(+) and Cl(-). These are called spectator ions because they do not participate in the reaction. For the net ionic equation they are left out since because they are on both sides of the equation they cancel out.

Net ionic equation:

NH4(+)(aq) + OH(-)(aq) ----> NH3(g) + H2O(l)

4 Balanced:
NaCH3COO(aq) + HCl(aq) --> NaCl(aq) + CH3COOH(aq)


Complete Ionic:
Na+ +CH3COO- + H+ + Cl- --> Na+ + Cl- + CH3COO- + H+

There is no net ionic equation because all of the ions cancel out--there is no reaction because all of the products are soluble in water.

5 Molecular equation

Na2S(aq) + 2HCl(aq) ------------> 2NaCl(aq) + H2S(g)

Then, write out the TOTAL IONIC EQUATION
Do this by seperating anything in aqueous solution into its constituant ions...

2Na+(aq) + S2-(aq) + 2H+(aq) + 2Cl-(aq) -----> 2Na+(aq) + 2Cl-(aq) + H2S(g)

Now, you can see that 2Na+(aq) and 2Cl-(aq) occur on both sides of the equation unchanged, they are called spectator ions and do not participate in the reaction.

Spectator ions are left out of the

NET IONIC EQUATION

2H+(aq) + S2-(aq) ----------> H2S(g)

6 The molecular equation is
Pb(NO3)2 (aq)+ Na2S (aq) >> 2 NaNO3 (aq)+ PbS (s)

ionic equation
Pb2+ (aq) + 2 NO3-(aq) + 2 Na+ (aq) + S2-(aq) >>
2Na+ (aq) + 2 NO3-(aq) + PbS (s)

the net ionic equation is
Pb2+ (aq) + S2- (aq) >> PbS (s)

7 Molecular -- combine the ionic compounds back

Pb(NO3)2(aq) + H2SO4(aq) = PbSO4(s) + 2 HNO3(aq)

Ionic

Pb(2+) + 2 NO3(-) + 2 H(+) + SO4(2-) = PbSO4(s) + 2 H(+) + 2 NO3(-)

All ionic species are (aq)

Net ionic -- cancel the spectator ions on both sides

Pb(2+)(aq) + SO4(2-)(aq) = PbSO4(s)

8 The molecular equation is

KCl + NaNO3 --> KNO3 + NaCl


The complete ionic equation is

K+ + Cl- + Na+ + NO3- --> K+ + NO3- + Na+ + Cl-

As for the net ionic equation, I'd say we can't write one because all of the ions here are spectator ions. They all appear on both sides so they all cancel out. This shows that both products are soluble in water.

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