Posted by Andrew on Sunday, July 14, 2013 at 1:35pm.
Why are you stuck. The procedure is correct and your answer is correct; however, if you are concerned about the number of significant figures, you are allowed only two (from the 8.0); therefore, that 295.07grams should be rounded to 3.0E2 grams.
Thank you so much! Now for the second part of the question. How many ml of that stock solution would you need to prepare 175 ml of 0.750 M diluted Na3PO4 solution? My answer is 4.27 ml of stock solution. Is that correct? Thank you. If yes , the next part of the question is: if you added 3.2652 g of CaCl2 to your diluted Na3PO4 solution, what is the limiting reactant?
No.
The dilution formula is
c1v1 = c2v2 where
c = concn
v = volume
8*v1 = 0.750*175
Solve for v1 = approximately 16 mL.
3CaCl2 + 2Na3PO4 ==> Ca3(PO4)2 + 6NaCl
mols CaCl2 = grams/molar mass = estimated 0.002
mols Na3PO4 = M x L = estimated 0.0012.
Now convert mols Na3PO4 and mols CaCl2 to mols Ca3(PO4)2 using the coefficients in the balanced equation. .
Estimated numbers are as follows:
0.002 mols CaCl2 x (1 mol Ca3(PO4)2/3 mols CaCl2) = 0.002*1/3 = est 0.00067 mols Ca3(PO4)2 solid.
0.0012 mols Na3PO4 x (1 mol Ca3(PO4)2/2 mol Na3PO4) = 0.0012 x 1/2 -= est 0.0006
Note that the two values are different which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent; therefore, Na3PO4 solution is the limiting reagent. That means that some of the CaCl2 will not react.
Thank you for your time! Obviously I need to review the equation set-ups again. You were very helpful!