# Chemistry

Determine the mass of PbSO4 produced when 10mL of 0.2M CuSO4 reacts with 10mL of 0.2M Pb(NO3)2?

Balance the equation
CuSO4 + Pb(NO3)2 → PbSO4 + Cu(NO3)2

Find the number of moles for CuSO4
c=n/v
n=c × v
n=0.2 × 0.01
n=0.002 mol
Since it is a 1:1 ratio, the number of moles for Pb(NO3)2 is the same as that for CuSO4.
Therefore, there is no limiting reagent.

Use the moles of CuSO4 to find the moles of PbSO4
(Things that are known)/(Things that are unknown) = 1/1 = 0.002/x

1 × x = 1 × 0.002
Therefore, x = 0.002mol of PbSO4 is the theoretical yield

Find molar mass of PbSO4
M(PbSO4) = 207.2 + 32 + (16×4)
=303g/mol

Now find the mass of PbSO4 using m = nM
m=nM
m=0.002 × 303
m=0.606g of PbSO4 is the predicted weight of the precipitate

Find the molar mass of CuSO4.5H2O
M(CuSO4) = 63.546 + 32.06 + (16 × 4)
= 159.61
M(5H2O) = (10 × 1) + (5 × 16)
= 90
M(CuSO4.5H2O) = 159.61 + 90
= 249.61g/mol

Find number of moles for CuSO4.5H2O
c=n/v
n=c × v
n=0.2 × 0.1
n=0.02 mol

Now find the mass of CuSO4.5H2O using m = nM
m=nM
m=0.02 × 249.61
m=4.99g of CuSO4.5H2O is needed to make a 100mL 0.2M solution

Find the molar mass of Pb(NO3)2
M(Pb(NO3)2) = 207.2 + (2 × 14.007) + (6 × 16)
= 331.208g/mol

Find number of moles for Pb(NO3)2
c=n/v
n=c × v
n=0.2 × 0.1
n=0.02 mol

Now find the mass of Pb(NO3)2 using m = nM
m=nM
m=0.02 × 331.208
m=6.62g of Pb(NO3)2 is needed to make a 100mL 0.2M solution

This all of this correct?

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1. Yes and no. I have copied the problem below and added my comments in bold.

Balance the equation
CuSO4 + Pb(NO3)2 → PbSO4 + Cu(NO3)2

Find the number of moles for CuSO4
c=n/v
n=c × v
n=0.2 × 0.01
n=0.002 mol
Since it is a 1:1 ratio, the number of moles for Pb(NO3)2 is the same as that for CuSO4.
Therefore, there is no limiting reagent.
Down to here you have done excellent work.

Use the moles of CuSO4 to find the moles of PbSO4
(Things that are known)/(Things that are unknown) = 1/1 = 0.002/x

1 × x = 1 × 0.002
Therefore, x = 0.002mol of PbSO4 is the theoretical yield
Since I want PbSO4 I would have used mols Pb(NO3)2; however, since they react exactly and there is no limiting reagent I suppose it actually makes no difference which you use. Again, good work.

Find molar mass of PbSO4
M(PbSO4) = 207.2 + 32 + (16×4)
=303g/mol
Since you used 4 places for Pb (207.2) I would have used 4 places for the answer and used 303.3 for the molar mass of PbSO4. I don't think this is a big deal but some profs will count off for that.

Now find the mass of PbSO4 using m = nM
m=nM
m=0.002 × 303
m=0.606g of PbSO4 is the predicted weight of the precipitate
Very good except for the 303.3

I don't understand why you have done the rest of this to the end. The problem doesn't ask for CuSO4 (either concentration or grams) nor does it ask for Pb(NO3)2 or Cu(NO3)2. In addition you have complicated the problem by switching from CuSO4 to CuSO4.5H2O and back again. The original problem uses CuSO4 so any relevance to CuSO4.5H2O is missing (and unnecessary).

Find the molar mass of CuSO4.5H2O
M(CuSO4) = 63.546 + 32.06 + (16 × 4)
= 159.61
M(5H2O) = (10 × 1) + (5 × 16)
= 90
M(CuSO4.5H2O) = 159.61 + 90
= 249.61g/mol

Find number of moles for CuSO4.5H2O
c=n/v
n=c × v
n=0.2 × 0.1
n=0.02 mol

Now find the mass of CuSO4.5H2O using m = nM
m=nM
m=0.02 × 249.61
m=4.99g of CuSO4.5H2O is needed to make a 100mL 0.2M solution

Find the molar mass of Pb(NO3)2
M(Pb(NO3)2) = 207.2 + (2 × 14.007) + (6 × 16)
= 331.208g/mol

Find number of moles for Pb(NO3)2
c=n/v
n=c × v
n=0.2 × 0.1
n=0.02 mol

Now find the mass of Pb(NO3)2 using m = nM
m=nM
m=0.02 × 331.208
m=6.62g of Pb(NO3)2 is needed to make a 100mL 0.2M solution

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