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Simple stoichiometry questions

  1. Jun 29, 2013 #1
    Hey guys,

    I have two stoichiometry questions to do with solvents. Both of them are very simple and I am unable to complete them at the moment because of some silly error in my method, I am sure. Can you please explain the method and how to do the question in some detail? I really appreciate your help.

    6) What volume of 0.250M Sulphuric Acid (H2SO4) is needed to react completely with 13.5g of Sodium Hydroxide(NaOH)?

    According to this equation: 2NaOH(aq) + H2SO4 = Na2SO4(aq) + 2H2O(l)

    7) 200mL of 0.105 M silver nitrate is added drop wise to 25mL of sodium chloride until all the chloride precipitates as silver chloride:

    AgNO3(aq) + NaCl(aq) = AgCl(s) + NaNO3(aq)

    * What is the concentration of the original sodium chloride solution?

    * What is the mass of silver chloride precipitated?
  2. jcsd
  3. Jun 29, 2013 #2
    Here is a photo to clear any confusion (if any). If you could please explain question [8] too, it'll be great. Thanks.
  4. Jun 29, 2013 #3
    ImageUploadedByPhysics Forums1372504687.793743.jpg
  5. Jun 29, 2013 #4


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    Staff Emeritus
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    Homework Helper

    In order to receive help, you must show some work in solving these problems.
  6. Jun 29, 2013 #5
    Ok....so what's wrong with what I've done. ImageUploadedByPhysics Forums1372547888.980356.jpg
  7. Jun 30, 2013 #6


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    Staff: Mentor

    Concentration of NaCl is OK, but I have no idea what you did later and why. Volume of the NaCl solution was given, so there is no need to calculate it, and the question asked to calculate something else.
  8. Jun 30, 2013 #7
    Would this be right then?

    6) 2NaOH(aq) + H2SO4 = Na2SO4(aq) + 2H2O(l)
    n(NaOH) = m/M = 13.5/40 = 0.3375mol
    n(H2SO4) = n(NaOH) * 0.5 = 0.16875 mol
    n(H2SO4) = x L * 0.250molL^-1
    0.16875/0.250 = x = 0.675L

    0.675L of sulfuric acid is needed.

    7) AgNO3(aq) + NaCl(aq) = AgCl(s) + NaNO3(aq)
    n(AgNO3) = 0.105/5 = 0.021 mol
    n(NaCl) = n(AgNO3) = 0.021 mol
    n(NaCl) = 0.025L * y molL-1
    y = 0.021mol/ 0.025L = 0.84

    Concentration of NaCl solution = 0.84M

    n(AgCl) = n(NaCl) = 0.021mol
    m = nM = 0.021 * (107.8+35.5) = 3.01g

    Mass of precipitate = 3.01g
  9. Jun 30, 2013 #8


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    Staff: Mentor

    OK now.
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