Chemistry

posted by .

8.76g of aluminum sulphide is added to 0.0350L of 0.250mol/L lead(II)nitrate solution. calculate the maximum mass of precipitate, PbS that can form

  • Chemistry -

    Al2S3 + 3Pb(NO3)2 ==>3PbS + 2Al(NO3)3

    Convert 8.76 g Al2S3 to moles. moles = grams/molar mass
    Convert the Pb(NO3)2 to mols. moles = M x L.
    Now, using the coefficients in the balanced equation, convert each of the reactants to moles of PbS. Undoubtedly you will obtain different answers for moles PbS; the correct answer in limiting reagent problems like this is the smaller value. Then convert the smaller value to grams. g = moles x molar mass.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. CHEMISTRY

    a solution is made by diluting 175 mL of 0.238 M aluminum nitrate, Al(NO3)3, solution with water to a final vlume of 5.00x 10^2 mL.calculate the following. a. the molarity of aluminum nitrate, aluminum ion, and nitrate ion in the diluted …
  2. Chemistry Urgent Help

    if 4.55 g of sodium sulphide and 15.0 g of bismuth nitrate are dissolved in separate beakers of water which are then poured togehter, what is the maximum mass of bismuth sulphide an insoulble compound that could precipitate?
  3. Chemistry ....Please help

    If 4.55 g of sodium sulphide and 15.0 g of bismuth nitrate are dissolved in separate beakers of water which are then poured together, what is the maximum mass of bismuth sulphide, an insoluble compound, that could precipitate?
  4. College Chemistry

    1) The concentration of the aluminum ion in a test solution is 0.10M. Using the Ksp for aluminum hydroxide (3. x 10^-34), calculate the concentration of hydroxide needed to precipitate aluminum hydroxide. From the concentration of …
  5. Chemistry-Experiment

    I did an experiment where I had 5 ml of MgCl2, then 5 ml NH4 solution (NH4OH (aq)) was added and it was observed that a precipitate formed; cloudy solution. After I added 1g of NH4Cl(s), the solution clears up (transparent). 2nd experiment: …
  6. Chemistry

    15.0 mL of 0.30 M sodium phosphate solution reacts with 20.0 mL of 0.23M lead(II) nitrate solution. What mass of precipitate will form?
  7. Chemistry

    A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide solution. A bright yellow Precipitate is of lead(ii) iodide is formed. Calculate the mass of lead (ii) iodide that should be formed. …
  8. Chemistry

    We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. If we have 500. mL of 0.632 M lead(II) nitrate solution and add excess potassium iodide, how much solid lead(II) iodide …
  9. Chemistry

    I'm sorry here is another I do not understand. How do you find out exactly how much of a solution you would need to completely precipitate another compound in the reaction?
  10. Chemistry

    A chemist takes a 647 gram sample of this hydrate. She removes all the water possible from the sample and collects it. To this water, she adds 7.6 grams of lead(II) nitrate. Afterwards she wants to precipitate all the lead. She wants …

More Similar Questions