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?

  1. 👍
  2. 👎
  3. 👁
  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

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chem 11

    So I did a lab determining the chemical formula of a hydrate. I really just need somebody to check this over for me. We had to find the molecular formula of a hydrate of copper (2) sulphate, CuSO4 .xH2O. My observation table looks

  2. CHEMISTRY 2

    A 56.0mL sample of a 0.102 M potassium sulfate solution is mixed with 34.7mL of a 0.114 M lead acetate solution and this precipitation reaction occurs: K2SO4(aq)+Pb(C2H3O2)2(aq)→PbSO4(s)+2KC2H3O2(aq) The solid PbSO4 is

  3. science(chem)

    I synthesized Nylon in my lab for organic chem. I need help figuring out the theoretical yield so I can compare it to my product that I synthesized. I looked up the reaction online on Google by typing in "Synthesis of adipoyl

  4. chemistry

    The solubility product constant (Ksp) for the dissolution of PbSO4 as represented by the chemical equation is 1.7x10-8. PbSO4 (s)= Pb2+ (aq) + SO42- (aq) Calculate the mass (g) of PbSO4 that dissolves in 1100 mL of water.

  1. Chemistry 11

    So I did a lab determining the chemical formula of a hydrate. We had to find the molecular formula of a hydrate of copper (2) sulphate, CuSO4 .xH2O. My observation table looks like this: Mass of clean, dry test tube: 21.6g Mass of

  2. Chemistry

    I was doing a lab experiment on the reaction of copper(II) sulphate and aluminum: 1) I measured 2.02g of copper(II) sulphate pentahydrate 2) I dissolved the copper(II) sulphate pentahydrate in 10mL of distilled water 3) I added

  3. chemistry

    What is the balaced net ionic for this equation: PbNO3+CuSO4-->PbSO4+CuNO3

  4. Chemistry

    Given: PbSO4 + 2e− Pb + SO42− E° = -0.356 V PbO2 + 4H+ + SO42− + 2e− PbSO4 + 2H2O E° = 1.685 V Determine the cell voltage of the lead storage battery when the concentration of sulfuric acid is 4.00 M. Where do I start,

  1. chemistry

    A sample of a hydrate of CuSO4 with a mass of 250 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 160 grams. What is the formula for the hydrate? CuSO4 * 10 H2O CuSO4 * 7

  2. chemistry

    Calculate the mass of the solid you must measure out to prepare 100 mL of 0.025 M CuSO4. Note that this salt is a hydrate, so its formula is CuSO4⋅5H2O. You must include the waters of hydration when calculating the formula

  3. Organic Chemistry

    An impure sample contains 0.95g of benzoic acid and 0.05 of salicylic acid. Solubilities in water of the two compounds are given in the following table. Solubility at 20 degrees (g/10mL) benzoic acid: 0.029 salicylic acid: 0.22

  4. science

    how much salt must be added to how much water to create 10ml of a 10% saline solution.

You can view more similar questions or ask a new question.