Can a chemistry genius please help me with the following precipitation reactions promblem?

Question

I need to know how to do this. I'm going to be tested on this material.

Here is the problem:

Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write thechemical balanced equation, the overall ionic equation, and the net ionic equation for this reaction. If 13.45 g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?

I want to let everyone that any help you can give me is really good.

Answer 1

Chemistry genius to the rescue
heres how you do it
first you need to find the molar mass of copper II chloride, just add the atomic weight of chloride with 2 X atomic weight of chlorine
63.546+35.5+35.5=134.486
after that you have to find the number of moles of copper II chloride that reacts with lead II nitrate, to do this just divide the mass of copper II chloride that you have by the molar mass
13.45/134.486=.1 moles
now you need to write down the equation and balance it
CuCl2+Pb(NO3)2=PbCl2+Cu(NO3)2
this tells you that for every mole of copper II chloride you get 1 mole of each of the products
figure out which one is the precipitate, you should have your solubility rules memorized, nitrates are always soluble so PbCl2 is your precipitate
since you have .1 moles of CuCl2 you will have .1 moles of PbCl2
after that you need to find the molar mass of PbCl2, same way as with CuCl2
207.3+35.47+35.47=278.24
and finally multiply that by the number of moles of PbCl2 to get the mass of the precipitate
278.24*.1=27.83 grams



for overall ionic you need to split up everything but the precipitate up, put aq by everything and s by the precipitate
you get
Cu^(2+)(aq)+2Cl^(-)(aq)+
Pb^(2+)(aq)+2NO3^(-)(aq)=
PbCl2(s)+Cu^(2+)(aq)+
2NO3^(-)(aq)
to get the net ionic equation you cancel out the ions that are the same on both sides, the so called spectator ions they are called that because they don't actually participate in the reaction
in this case the Cus and the NO3s cancel
so your net ionic is
Pb^(2+)(aq)+2Cl^(-)(aq)=
PbCl2(s)

P.S. the ^(2+) means that the Pb has a charge of plus 2

Answer 2

You already asked the part about writing the equation so you can look at those so I won't repeat it again here.

lets assume a generic equation and I'll show you the setup. lets say A is CuCl2 and C is your precipitate

2A + B ----> C + 2D
formula mass for A = 134.5 (CuCl2)
formula mass for C = PbCl2 but I'll let you calculate it

?g ppt = (13.45g CuCl2) x
[( 1mole A)/(134.5g A)] x [(1 mol C)/(2 molA)] x
[(? g C)/(1 mol C)] = answer in g
where ? above is the value you calculated from the formula mass of PbCl2 above.

Answer 3

CuCl2 + Pb(NO3)2---> PbCl2 + Cu(NO3)2
The metals stay +2, the Cl stays -1