[john4614]

Hello, First time here.. Im looking for help with my problems i did them best to my knowledge not sure of all what it is asking.... Can someone please look over my work and tell me how i did and where i need to fix...Im really stuck in this subject:
A.   Consider a reaction in which 10.0ml 0.50M sodium iodide and 10.0ml of 0.25M lead(II)nitrate are both added into a 250ml beaker. The solution is gently heated to almost boiling and allowed to cool to room temperature. Write the balanced equation, identify the precipitate, identify the limiting reactant (if there is any) and predict the theoretical yield of the precipitate.

NaI: 0.50M x 0.01L=0.005mol
Ob(NO3)2: 0.25M x 0.001L= 0.0025mol
No reaction?

Balanced Molecular: Pb(NO3)2(aq) + 2NaI (aq)-> PbI(s) +2NaNO3 (aq)
Ionic: Pb(aq)+2NO3 (aq)+2Na(aq)+2I(aq)->PbI2(s)+2Na(aq)+2NO3(aq)
Net Ionic: Pb+2I->PbI2(s)

^^^^^^Is that answering the predict the theoretical yield of precipitate… is the answering identify the precipitate?

B. Reaction when 50.0ml of distilled is place in 250ml beaker and
Copper(II)sulfate 10.0ml of 0.25M
Barium 10.0ml of 0.25M
What to find: Balanced equation, identify the precipitate, identify the limiting reactant (if there is one) predict the theoretical yield of the precipitate… How did the addition of the distilled water affect your calculations?
CuSO4: 0.25M x 0.01L= 0.0025mol
BaCl2 0.25M x 0.001L= 0.0025mol
Both Reactants
CuSO4+BaCl2CuCl2+BaSO4
Cu+SO4+Ba+Cl2Cu+Cl2+Ba+SO4
Everything cancels out unless there is a solid im not finding?

C:
Equal volumes of 10.0ml solution of 0.25M magnesium chloride and 0.25M silver Nitrate place into a 250ml beaker
Find Balanced equation, net ionic, identify the precipitate and predict the theoretical yield of the precipitate
AgNo3:  0.25x0.01l=0.0025mol???
MgCl2:   0.25x0.01L=00.0025?

Balanced Molecular equation:
2AgNO3+MgCl22AgCl+Mg(NO3)2
Ionic:
Ag2(aq)+2NO3(aq)+Mg(aq)+Cl2(aq) 2AgCl(s)+Mg(aq)+2NO3(aq)
Net Ionic:
Ag2(aq)+Cl2(aq)2AgCl(s

 
D: Use equal volumes of 10.0ml solutions
Calcium Chloride= .25M
Sodium Carbonate= 0.25M
Write Balanced equation, net ionic equation, identify the precipitate and predict the theoretical yield of the precipitate.

CuCl2:  0.25x0.01l=0.0025mol???
NaCo3:   0.25x0.01L=00.0025?

Balanced Molecular:
CuCl2(aq)+2NaCo3(aq)- Ca(CO3)2(s)+2NaCl9aq)
Ionic:
Ca+Cl2+2Na+2CO3--Ca(CO3)2(s)+2Na+2Cl
Net Ionic:
Ca(aq)+2CO3(aq)-Ca(CO3)2(s)

[Astrokel]

A.

NaI: 0.50M x 0.01L=0.005mol
Ob(NO3)2: 0.25M x 0.001L= 0.0025mol
No reaction?

Not that there is no reaction, the stoichiometric ratio is satisfied, therefore there should not be any limiting reagent as the question ask.

Balanced Molecular: Pb(NO3)2(aq) + 2NaI (aq)-> PbI(s) +2NaNO3 (aq)
Ionic: Pb(aq)+2NO3 (aq)+2Na(aq)+2I(aq)->PbI2(s)+2Na(aq)+2NO3(aq)
Net Ionic: Pb+2I->PbI2(s)

typo: PbI₂

^^^^^^Is that answering the predict the theoretical yield of precipitate… is the answering identify the precipitate?

You are identifying the precipitae. Theoretical yield is in moles or grams.

B

Everything cancels out unless there is a solid im not finding?

Check again on the salt solubility.
C

Ag2(aq)+2NO3(aq)+Mg(aq)+Cl2(aq) 2AgCl(s)+Mg(aq)+2NO3(aq)
Net Ionic:
Ag2(aq)+Cl2(aq)2AgCl(s

2Cl^-
D

CuCl2(aq)+2NaCo3(aq)- Ca(CO3)2(s)+2NaCl9aq)

formula for calcium carbonate is wrong

[AWK]

Solubility of PbI₂ is about 3.2 x 10^-5 - it should be the precipitate.

[john4614]

Thanks

[john4614]

other than that are they all good? Am I answering what it ask? Why are they all .25M x 10.0ml=.25M x 0.01L? Do I have to do anything with the 250ml beaker ?

[Astrokel]

yes it looks good other than those pointed out(check D calculations).  Perhaps they are not testing you on the calculations but rather understanding on stoichiometry, so the numbers are the same. Don't worry about the beaker.