I need a chemistry genius! Will some1 please help me w/ Precipitation Reactions Problems? Need 2 no 4 test.?

Question

Please help me solve the following 2 problems. I need to understand how to do it.

1. Write the balanced chemical equation, write the overall ionic equation, identify the spectator ions and possible precipitates, and write the net ionic equation for each of the following reactions.
a. mercury(II) chloride (aq) + potassium sulfide (aq) ----->
b. sodium carbonate (aq) + calcium chloride (aq) ----->
c. copper(II) chloride (aq) + ammonium phosphate (aq) ----->

2. Identify the spectator ions in the reaction between KCl and
AgNO3 (3 = subscript) in an aqueous solution.

Thanks so much for anyone who is willing to help me.

Answer 1

a. HgCl2 + K2S ===> HgS + 2KCl
Spectator ions: K+, Cl-
Precipitate HgS
Net ionic equation: Hg2+ + S= ===> HgS

b. Na2CO3 + CaCl2 ===> CaCO3 + 2NaCl
Spectator ions: Na+, Cl-
Precipitate: CaCO3
Net ionic equation: Ca2+ + CO3= ===> CaCO3

c. 3CuCl2 + 2(NH4)3PO4 ===> Cu3(PO4)2 +6 NH4Cl
Spectator ions: NH4+, Cl-Precipitate: Cu3(PO4)2
Net ionic equation: 3Cu2+ + 2PO43- ===> Cu3(PO4)2

2. Ag+ NO3- + K+ + Cl- ===> AgCl Spectator ions: K+ NO3-

Answer 2

1) I'll explain one....

mercury(II) chloride (aq) + potassium sulfide (aq) ----->

identify the type of reaction and since you know there is a precipitate, it is double displacement so the ions will exchange places to form

mercury(II)sulfide(s)+ potassium chloride(s)

Check your solubility table and you'll see that Hg(II)S is the solid or precipitate and KCl so it will remain in ionic form or dissolved.

so we have
Hg(II)Cl₂(aq) + K₂S(aq) ---> HgS(s) + 2KCl(aq)
Hg(II) tells you the charge is +2 so you need 2 Cl- ions to balance the charge. The charge on the sulfide ion is -2 so it balances the Hg ion in the products.

the 2 in front of KCl is to keep the equation balanced.

Now, to get the ionic eq, write all the soluble substances as ions and solids as compounds. I'll use ^ for all superscipts
Hg^2+(aq) + 2Cl^-(aq) +2K^+(aq) + S^2-(aq) --->
HgS(s) + 2K^+(aq) + 2Cl^-(aq)

subtract out identical particles on each side(meaning free floating ions) to get the net equation

Hg^2+(aq) + S^2-(aq) ---> HgS(s)

the ions the same on both sides that you just removed are the spectator ions.

For #2, you'll need to do the same thing as in a,b&c but they are asking for the ions not part of the net ionic equation. In the example just completed,
2K^+(aq) + 2Cl^-(aq) are the spectator ions.

Answer 3

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