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Simple Stioch

  1. Jun 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok so I should know how to do this but I want to make sure.
    The balanced equation involved is:
    [tex]2Na_{3}PO_{4}*12H_{2}0+3BaCl_{2}*2H_{2}0\longrightarrow Ba_{3}(PO_{4})_{2}+6NaCl+30H_{2}0[/tex]
    In the lab I obtained 6.6e-4 moles of [tex]Ba_{3}(PO_{4})_{2}[/tex]
    and [tex]2Na_{3}PO_{4}*12H_{2}0[/tex] is limiting we found out.
    I just need to convert moles of [tex]Ba_{3}(PO_{4})_{2}[/tex] to grams
    [tex]2Na_{3}PO_{4}*12H_{2}0[/tex]. I know how to convert from moles grams etc. but i cant decide when i do the mole to mole ratio. do I use 2*12=8moles for the [tex]2Na_{3}PO_{4}*12H_{2}0[/tex] or just 2?
    Thanks

    2. Relevant equations
    [tex]2Na_{3}PO_{4}*12H_{2}0+3BaCl_{2}*2H_{2}0\longrightarrow Ba_{3}(PO_{4})_{2}+6NaCl+30H_{2}0[/tex]


    3. The attempt at a solution

    I know how to convert from moles grams etc. but i cant decide when i do the mole to mole ratio. do I use 2*12=8moles for the [tex]2Na_{3}PO_{4}*12H_{2}0[/tex] or just 2?
    Thanks
     
  2. jcsd
  3. Jun 16, 2007 #2

    symbolipoint

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    1 mole of barium phosphate required 2 moles of sodium phosphate:12 hydrate. Straightforward stoichiometry. Arrange your whole conversion expression and include the units for each number. Start with the 1 mole of barium phosphate!

    (1 mole ba.phos)*[(2 mole sod.phos.hydrate)/(1 mole ba.phos)]*(U grams/mole of sod.phos.hydrate)

    You will need to write that all conventionally with pencil on paper; I used the symbol "U" for the number of grams per mole for sodium phosphate 12 hydrate, but you need to find it in a table or calculate it yourself.
     
  4. Jun 16, 2007 #3
    Yah know how to set it up, it just like i said before do I multiply with 24 or 2. Anyways I got 0.50g of [tex]2Na_{3}PO_{4}*12H_{2}0[/tex]
    and
    0.48g of [tex]BaCl_{2}*2H_{2}0[/tex]
     
  5. Jun 16, 2007 #4

    symbolipoint

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    You need to know what the numbers in the reaction statements mean. If you set up the calculation correctly, such as I have shown, then you will obtain the correct result. The two things applied in my setu-up expression which were not done were full-correct names for the compound (I used my own abbreviations) and the "U" in place of the actual formula weight.
     
  6. Jun 16, 2007 #5
    yah so:
    [tex] (6.6e^-4 moles Ba_{3}(PO_{4})_{2}) *\frac{2BaCl_{2}*12H_{2}O}{1 mole Ba_{3}(PO_{4})_{2}}*\frac{380.12 2BaCl_{2}*12H_{2}O}{1 mole BaCl_{2}*12H_{2}O} [/tex]

    I get 0.50 right?
     
    Last edited: Jun 16, 2007
  7. Jun 16, 2007 #6
    right!!!!???
     
  8. Jun 17, 2007 #7

    symbolipoint

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    The only problem I see in the arranged expression is the formula weight is labeled as barium chloride - but you are not interested in that. You are interested in the sodium phosphate 12*hydrate. The the formula weight is the correct one for the phosphate.
     
  9. Jun 17, 2007 #8
    Fix it to
    [tex] (6.6e^-4 moles Ba_{3}(PO_{4})_{2}) *\frac{2Na_{3}PO_{4}*12H_{2}0}{1 mole Ba_{3}(PO_{4})_{2}}*\frac{380.12 2Na_{3}PO_{4}*12H_{2}0}{2Na_{3}PO_{4}*12H_{2}0} [/tex]
     
    Last edited: Jun 17, 2007
  10. Jun 17, 2007 #9
    :grumpy:So is 0.50g right!?
     
  11. Jun 17, 2007 #10
  12. Jun 17, 2007 #11

    symbolipoint

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    YES. Probably closer to about 0.502 g. I hope this helps.
    Why were you unsure?
     
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