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General : Salt Mixture Composition ( Limiting Reactant)
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 Message 1 of 5 in Discussion 
From: MSN NicknameN_2006  (Original Message)Sent: 1/29/2008 3:46 AM
Hey Steve,
I have naother question for you, thanks for helping me out with the last one.
We did a lab and now I am writing up the post lab report and I am having some trouble with figuring out the salt mixture concentrations.
It asks to determine the percentage of BaCl2 H2O   and percentage of Na3PO4 12H2O
 
So I have the mass and the moles of Ba3(PO4)2
 
How do I go about doing this?   Don't I need more information to calculate the answer?

Thanks in advance!


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Reply
 Message 2 of 5 in Discussion 
From: MSN Nickname·Steve·Sent: 1/29/2008 6:06 AM
All you need to know is which reactant, Na3PO4·12H2O or BaCl2·2H2O, is the limiting reactant by testing the supernatant from the solid Ba3(PO4)2 product.  If you add a few drops of Na3PO4 solution to the supernatant and get a precipitate, Na3PO4·12H2O is the limiting reactant and BaCl2·2H2O is in excess.  If you add a few drops of BaCl2 to the supernatant and get a precipitate, BaCl2·2H2O is the limiting reactant and Na3PO4·12H2O is in excess.

Reaction for reference:
2 Na3PO4·12H2O (aq)  +  3 BaCl2·2H2O (aq)   ––â€?gt;   Ba3(PO4)2 (s)  +  6 NaCl (aq)  +  30 H2O (l)

From the moles of Ba3(PO4)2 you got, calculate the moles of the limiting reactant according to the balanced chemical reaction, and then convert moles of the limiting reactant to grams by multiplying the moles by its formula weight (include the waters of hydration in the formula weight).  Subtract the weight of the limiting reactant from the original mixture weight to get the weight in grams of the other reactant in the original mixture.

Percent BaCl2·2H2O    =    grams of BaCl2·2H2O     X   100
                                        total grams of mixture

Percent Na3PO4·12H2O    =    grams of Na3PO4·12H2O    X   100
                                               total grams of mixture

Reply
 Message 3 of 5 in Discussion 
From: MSN NicknameN_2006Sent: 1/29/2008 8:02 PM
ALright.. thanks again!
 
So the limiting reactant for this experiment was the BaCl2
ANd we measured the salt mixture to be 0.9844 g and the mass of precpitate of Ba3(PO4)2 was 0.2267g
 
SO you said to take the moles of limiting reactant and multiply by the molecular weight, so would I take the moles of Ba3(PO4)2 and use the stoich. ratio, which is 3/1 so then multi the moles of Ba3(PO4)2 to get moles of BaCl2
 
So the moles of Ba3(PO4)2 were calculated to be 0.0003766 moles. Then I multiply by 3/1 which is then 0.0011298 moles of BaCl2
then take molar mass of formula weight and multiply to get the mass of  grams of BaCl2·2H2O  which then I would I subtract from the total, which was 0.9844 g to get the Sodium phosphate mass, which then I can calculate the percentages.
 
I hope I am right. I read what you wrote and then I just wanted to make sure I am doing the calculations correctly.
Thanks for your help!!!!

Reply
 Message 4 of 5 in Discussion 
From: MSN Nickname·Steve·Sent: 1/29/2008 8:34 PM
Yes, that is right.

Reply
 Message 5 of 5 in Discussion 
From: MSN Nickname·Steve·Sent: 1/29/2008 8:46 PM
It's kind of a clever experiment, and gives lots of practice adding up those big formula weights! 

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