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Finding percent yield of Alum?

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the percent yield of alum

    In an experiment to synthesize alum I started with 0.766g of aluminum and made 8.246g of alum

    2. Relevant equations
    2Al + 6H20 + 2KOH ---> 2KAl(OH)4 + 3H2

    3. The attempt at a solution

    First I find the LR. Which in this expirment is Al.

    I find the moles of aluminum to be 0.02838mol, next I multiply by the mole ratio and by the molar mass of alum to find the theoretical yield of alum.

    Now to find the percent yield I do actual/theoretical X100%

    8.246g/3.806g x 100% = 217%. I know Im wrong because this is impossible.

    What did I do wrong?
  2. jcsd
  3. Sep 24, 2013 #2


    User Avatar

    Staff: Mentor

    Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.
  4. Sep 24, 2013 #3
    Well we also recieved another formula in this lab:

    KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

    Should I add this to the formula I listed in my first post to get a overall reaction equation?
  5. Sep 24, 2013 #4
    Hmm I still get a yield more than 100% after doing this. Which is physically impossible
  6. Sep 24, 2013 #5

    Is it allowed if I write the formula form the first equation except replace KAl(OH) for KAl(SO4)2
  7. Sep 24, 2013 #6
    Also, even if I use the formula u refer to I get a percent yield of 112% which is physically impossible in every sense and wrong.
  8. Sep 24, 2013 #7
    The formula is still wrong. Search for it on google.
  9. Sep 24, 2013 #8
    I get 61.1% yield. is this correct?
  10. Sep 25, 2013 #9
    I got this number using the fomula:

    2Al +2KOH +4H2SO4 + 22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O

    This was the only formula that helped produce a reasonable percent yield (61.1%). I am highly sceptical that there is an alternative to this.
  11. Sep 25, 2013 #10


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    Staff: Mentor

    That looks OK to me.
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