1 is a stoichiometry problem; I expect you've worked many of those. Here is a worked example to show how you convert moles Na2C2O4 (moles = M x L) to moles La2(C2O4)3.
http://www.jiskha.com/science/chemistry/stoichiometry.html
For #2, note that millimoles LaCl3 = M x mL = 25 x 0.15 = 3.75
mmoles Na2C2O4 = 30 x 0.2 = 6.
Write and balance the equation, then set up an ICE chart.
..2LaCl3 + 3Na2C2O4>La2(C2O4)3 + 6NaCl
I..3.75.....6.........0...........0
Using the conversions you used in part 1, convert moles LaCl3 to moles La2(C2O4) and do the same for Na2C2O4 to the La ppt. You will have two different answers and both can't be right. The correct answer, in limiting reagent problems (that's what you are determining at this point) is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. This will give you the same number of moles of the ppt formed as you determined in part 1. Now convert the moles of the limiting reagent to moles of the "other" reagent to see how much is used. Subtract from the initial amount to determine the excess of the other reagent. Go through in like manner for the other product, then each of the ions remaining in soln is M = millimoles/mL. Total mL = 25 + 30 = 55 mL.
Here is an example of a simpler equation and how that is handled.
20 mL 0.1M NaCl + 30 mL 0.1 AgNO3 and you can see NaCl is the limiting reagent without a lot of calculation.
......NaCl + AgNO3 ==> AgCl(s) + NaNO3
init....2......3..........0......0
change..-2.....-2.........2......2
equil...0.......1.........2......2
So you have 2 millimoles AgCl and moles x molar mass = grams AgCl.
Na^+ = 2 mmoles/50 mL = M
NO3^- = 3 mmoles/50 mL = M
M Ag^+ = 1 mmol/50 mL
M Cl^- = M of a saturated solution of AgCl which you can calculate from the Ksp.
When a solution of sodium oxalate (Na2C2O4) is added to a solution of lanthanum (III) chloride, lanthanum oxalate (molar mass 541.86 g/mol) The balanced net ionic equation for the reaction is:
3C2O4^2- (aq) + 2La^3+ (aq) --> La2(C2O4)3 (s)
Thirty ml of 0.2000 M sodium oxalate are added to 25.00 ml of 0.1500 M La Cl3.
1) How many grams of La2(C2O4)3 (s) are obtained, assuming 100% yeild?
2) What are the concentrations (M) of La^3+, C2O4^2-, Na^+, and Cl^- after the reaction is complete? (Assume that the volumes are additive.)
Please help! Number one makes some sense but I don't understand 2 at all!
5 answers
I'm trying to help my daugther with the SAME problem....your description above allowed us to recognize our mistakes, but we are still unsure if we are doing it right. Is there any way you can solve the above problem so we can follow the steps?
Thank you that was an extremely helpful explanation, but what is an ICE chart? I've never heard of that before
For Jean: I solved number one and got 0.47 g of La2(C2O4)3, so you can check your daughters answer. (Of course I'm not 100% sure but #1 is a lot easier than #2 which I'm still trying to figure out)
ICE chart.
I = initial
C = change
E = equilibrium.
2LaCl3 + 3Na2C2O4>La2(C2O4)3 + 6NaCl
mmoles Na2C2O4 = mL x M = 30 x 0.2M = 6
mmoles LaCl3 = mL x M = 25 x 0.15 = 3.75
Convert mmoles each to mmol product.
6 x [1 mol La2C2O4)3/3 mol Na2C2O4] = 2 mmoles product.
3.75 x [1 mole La2C2O4)2/2 mol LaCl3] = 1.875 mmoles product.
Therefore LaCl3 is the limiting reagent and 1.875 mmole product will be formed. That (0.001875 mole x 541.86 = 1.0159 which rounds to 1.016 g La2(C2O4)3.
Here is what we have.
LaCl3 is the limiting reagent. 3.75 mmol LaCl3 will use how much Na2C2O4? That will be 3.75 x (3 mol Na2C2O4/2 mol LaCl2) = 3.75 x (3/2) = 5.625 mmoles Na2C2O4. I can put all of that into an ICE chart as follows:
...2LaCl3 + 3Na2C2O4>La2(C2O4)3 + 6NaCl
I...3.75.....6.........0...........0
C..-3.75....-5.625...1.875......11.25
E...0.......0.375....1.875......11.25
Now you can use the equilibrium amounts to focus on the calculations. Remember to change mmole in the ICE chart to moles if you use M = moles/L or use M = mmol/mL.
Na^+ = Cl^- = 11.25 mmol/55 mL = ?
C2O4^2- = if we consider the La2(C2O4)3 to be insoluble and not produce ANY La^3+ or C2O4^2- ions (which isn't exactly right but I suspect this is what your teacher wants), then oxalate ion just that amount in excess since it was not the limiting reagent = (0.375 mmol/55 mL) = ?
IF you teacher wants you go through the solubility product, that's another problem. I hope this helps. I will post this at the top of the page today (late late Friday).
I = initial
C = change
E = equilibrium.
2LaCl3 + 3Na2C2O4>La2(C2O4)3 + 6NaCl
mmoles Na2C2O4 = mL x M = 30 x 0.2M = 6
mmoles LaCl3 = mL x M = 25 x 0.15 = 3.75
Convert mmoles each to mmol product.
6 x [1 mol La2C2O4)3/3 mol Na2C2O4] = 2 mmoles product.
3.75 x [1 mole La2C2O4)2/2 mol LaCl3] = 1.875 mmoles product.
Therefore LaCl3 is the limiting reagent and 1.875 mmole product will be formed. That (0.001875 mole x 541.86 = 1.0159 which rounds to 1.016 g La2(C2O4)3.
Here is what we have.
LaCl3 is the limiting reagent. 3.75 mmol LaCl3 will use how much Na2C2O4? That will be 3.75 x (3 mol Na2C2O4/2 mol LaCl2) = 3.75 x (3/2) = 5.625 mmoles Na2C2O4. I can put all of that into an ICE chart as follows:
...2LaCl3 + 3Na2C2O4>La2(C2O4)3 + 6NaCl
I...3.75.....6.........0...........0
C..-3.75....-5.625...1.875......11.25
E...0.......0.375....1.875......11.25
Now you can use the equilibrium amounts to focus on the calculations. Remember to change mmole in the ICE chart to moles if you use M = moles/L or use M = mmol/mL.
Na^+ = Cl^- = 11.25 mmol/55 mL = ?
C2O4^2- = if we consider the La2(C2O4)3 to be insoluble and not produce ANY La^3+ or C2O4^2- ions (which isn't exactly right but I suspect this is what your teacher wants), then oxalate ion just that amount in excess since it was not the limiting reagent = (0.375 mmol/55 mL) = ?
IF you teacher wants you go through the solubility product, that's another problem. I hope this helps. I will post this at the top of the page today (late late Friday).
1 answer