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anonymous

  • one year ago

If 4.88 grams of Zn reacts with 5.03 grams of S8 to produce 6.02 grams of ZnS, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem. unbalanced equation: Zn + S8 yields ZnS Someone please help me!!!

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  1. JFraser
    • one year ago
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    first and most important, balance the equation

  2. JoannaBlackwelder
    • one year ago
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    Yep, @JFraser is right. Do you know how to balance, @firesquad ?

  3. anonymous
    • one year ago
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    Balanced equation: 8Zn + S8 ---> 8ZnS

  4. anonymous
    • one year ago
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    @JFraser @JoannaBlackwelder

  5. JoannaBlackwelder
    • one year ago
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    Yep, good. Percent yield = (actual/theoretical)*100

  6. JoannaBlackwelder
    • one year ago
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    Actual is the amount of product given in the problem.

  7. JoannaBlackwelder
    • one year ago
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    Theoretical is the amount of product found using stoichiometry from the amount of reactant given.

  8. JoannaBlackwelder
    • one year ago
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    *from the amount of limiting reactant

  9. anonymous
    • one year ago
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    I am stuck right here 4.88 g Zn x (1 mol Zn/ 65.38 g Zn) = 0.07464 mol Zn 5.03 g S8 x (1 mol S8/ 256.48 g S8) = 0.01961 mol S8

  10. JoannaBlackwelder
    • one year ago
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    Ok, now use the molar ratios from the balanced reaction to convert to moles of ZnS

  11. anonymous
    • one year ago
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    what do i do after

  12. JoannaBlackwelder
    • one year ago
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    For both calculations.

  13. anonymous
    • one year ago
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    would the ratio be 1: 8

  14. JoannaBlackwelder
    • one year ago
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    For S8:ZnS, yes

  15. JoannaBlackwelder
    • one year ago
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    But not for Zn:ZnS

  16. anonymous
    • one year ago
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    0.07464 mol Zn * (1/1)= 0.07464 mol Zn 0.01961 mol S8 * (8/1) = 0.1569 mol S8 is that how to do it

  17. JoannaBlackwelder
    • one year ago
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    Well, that looks good for everything but the units at the end.

  18. anonymous
    • one year ago
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    What do I have to change for the units?

  19. JoannaBlackwelder
    • one year ago
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    0.07464 mol ZnS 0.1569 mol ZnS

  20. JoannaBlackwelder
    • one year ago
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    See why I did that?

  21. anonymous
    • one year ago
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    Oh Ok, now I find the percent yield right?

  22. JoannaBlackwelder
    • one year ago
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    Well, we need to first determine which value of product to go with.

  23. JoannaBlackwelder
    • one year ago
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    This is the limiting reactant part. Do you know how to determine that?

  24. anonymous
    • one year ago
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    The limiting reactant is S8

  25. JoannaBlackwelder
    • one year ago
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    No, the limiting reactant is the reactant that produces less product.

  26. anonymous
    • one year ago
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    Sorry, it is Zn

  27. JoannaBlackwelder
    • one year ago
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    Right :-)

  28. JoannaBlackwelder
    • one year ago
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    So, what is the mass of product using the limiting reactant?

  29. anonymous
    • one year ago
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    Sorry for wait I had to do something. It is 0.07464 mol ZnS

  30. JoannaBlackwelder
    • one year ago
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    No worries. That is the moles of product. What is the mass of product?

  31. anonymous
    • one year ago
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    65.38 g Zn

  32. JoannaBlackwelder
    • one year ago
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    Hm, that's not what I get. How did you do that?

  33. anonymous
    • one year ago
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    I got it from the first formula

  34. JoannaBlackwelder
    • one year ago
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    Hm, let me show you how I would do this step: 0.07464 mol ZnS(97.44 g/1 mol) = ? g ZnS

  35. anonymous
    • one year ago
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    Ok I get it

  36. JoannaBlackwelder
    • one year ago
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    :-) So, can you find the percent yield?

  37. anonymous
    • one year ago
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    Wait 7.27 is the theoretical yield

  38. JoannaBlackwelder
    • one year ago
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    Yep

  39. anonymous
    • one year ago
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    so what is the actual yield?

  40. JoannaBlackwelder
    • one year ago
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    The amount of product given in the problem.

  41. anonymous
    • one year ago
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    Thank you so much

  42. JoannaBlackwelder
    • one year ago
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    You're welcome! :-)

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