Chemistry 219

posted by .

An amount of 1.250g of an alloy (90% Ag, 10% Cu) was treated with HNO3 and H2SO4 to form Ag2SO4 and CuSO4. The solution was then treated with Cu wire to precipitate silver. The resulting solution was treated with excess zinc to precipitate copper. (show formula)
a). What mass of copper will be precipitated?
b). what total mass of zinc will be dissolved (neglect any dissolved by excess H2SO4)? Avoid any unnecessary steps in solving this problem.

thank you!!

  • Chemistry 219 -

    1.250g x 0.90 = ??g Ag
    1.250 x 0.10 = ??g Cu
    __
    Ag| + HNO3 ==> Ag2SO4
    Cu| + HNO3 ==> CuSO4

    2Ag^+ + Cu(s)(wire) = Ag(s) + Cu^+2(aq)
    Then Zn(s) + Cu^+2(aq) ==> Zn^+2(aq) +
    Cu(s)

    You know grams Ag initially. Convert to moles and use stoichiometry to calculate the amount of Cu(II) added to the solution when Ag was pptd. Here is a stoichiometry problem if you need it to get started.
    http://www.jiskha.com/science/chemistry/stoichiometry.html
    Then Cu initially + Cu added to ppt Ag, gives moles Cu reacting with Zn.
    Stoichiometry will give you the moles of Zn which can be converted to grams Zn.
    In addition to neglecting H2SO4 we must also assume there is no excess of HNO3.
    __

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chem- follow up question

    Sorry i posted this as an answer after my question had been answered, this is continued from the qualitative analysis question: That reallly cleared some things up, but I do still have a few questions. Umm im not completely clear on …
  2. Gen Chemistry

    52.8 mL of a calcium chloride solution with unknown concentration was treated with phosphoric acid to remove all of the calcium ions in the form of a precipitate. The precipitate was filtered, dried and was found to have the mass of …
  3. Chem 219

    If 48.9g of barium chloride dihydrate, BaCl2.2H20, was treated with excess silver nitrate solution. (show formula) a). how many moles of silver chloride precipitated?
  4. chemestry

    a coin is composed of gold and silver. the 5.00g coin is dissolved in nitric acid and the solution treated with NaCl to precipitate out 2.35g of AgCl. Gold chlorides do not precipitate under the same conditions. What is the composition …
  5. chemistry

    a portion (10.0 ml)of a solution containing the chloride anion was treated with silver nitrate to yield a precipitate (0.1713 g) of the silver chloride. calculate the molar concentration of the chloride anion in the solution. first …
  6. Chemistry

    A solution containing a mixture of metal cations was treated with dilute HCl and no precipitate formed. Next, H2S was bubbled through the acidic solution. A precipitate formed and was filtered off. Then, the pH was raised to about …
  7. Chem Help

    Consider a solution that contains Ag+ , Ba2+, and Pb2+ each at a concentration of .2M a) You add NaCl until the concentration of Cl- is 5.0 x 10^-3 M. A white precipitate forms. How do you determine whether that precipitate was AgCl …
  8. Chemistry need help please!!!!

    Write the formula of the anion present (i.e. anions are charged; do not write compounds) in these unknowns as indicated by these test results. 1) An unknown solution was treated with ammonium hydroxide and then with barium chloride. …
  9. chemistry

    . A student added 40.0 mL of an NaOH solution to 90.0 mL of 0.400 M HCl. The solution was then treated with an excess of nickel(II) nitrate, resulting in the formation of 1.06 g of Ni(OH)2 precipitate. Determine the concentration of …
  10. chemistry

    . A student added 40.0 mL of an NaOH solution to 90.0 mL of 0.400 M HCl. The solution was then treated with an excess of nickel(II) nitrate, resulting in the formation of 1.06 g of Ni(OH)2 precipitate. Determine the concentration of …

More Similar Questions