Posted by Jon on Sunday, March 13, 2011 at 10:47pm.
CaCl2 + 2AgNO3 ==> Ca(NO3)2 + 2AgCl
moles CaCl2 = M x L = ??
moles AgNO3 = M x L = ??
Go through the stoichiometry based on the coefficients in the balanced equation to see which reactant has some remaining unreacted. Post your work if you get stuck.
2AgNO3+CaCl2===>2AgCl+Ca(NO3)2 is the balanced equation.
so 0.2 x 0.1 = 0.2mol
= 0.2 (107.9*35.5)= 28.7g
0.15 x 0.1 = 0.015mol
0.015 (110.98)= 1.6647
??
or is it 0.20M x 0.1= 0.02M AgCl
0.15 x 0.01= 0.015M Cl
Neither. You corrected one error from the first post to the second but not the other.
moles AgNO3 = 0.2M x 0.1L = 0.02 moles. Convert to moles AgCl using the coefficients in the balanced equation.
0.02 moles AgNO3 x (2 moles AgCl/2 moles AgNO3) = 0.02 x (2/2) = 0.02 moles AgCl. No need to convert to grams.
moles CaCl2 = 0.15M x 0.1L = 0.015 moles.
Convert to moles AgCl. 0.015 moles CaCl2 x (2 moles AgCl/1 mole CaCl2) = 0.015 x (2/1) = 0.03 moles AgCl.
In limiting reagent problems like this the correct answer is ALWAYS the smaller value and the reagent producing that value is called the limiting reagent. So AgNO3 is the limiting reagent, we obtain 0.02 moles AgCl as the theoretical yield and we have how much CaCl2 that didn't react? I prefer to break this up into ions at this point to keep from getting mixed up.
CaCl2 we have 0.015 mole Ca+2 and 0.03 mole Cl-.
AgNO3 we have 0.02 mole Ag+ and 0.02 mole NO3-.
So all of the Ca+2 is left intact; M = 0.015moles/0.200L = ??M
Cl- = 0.030-0.02 = 0.01 and that divided by 0.2 L.
Ag+ is gone (but you CAN calculate the solubility of AgCl from Ksp).
NO3- = unchanged from original = 0.02M BUT it has been diluted from 100 mL to 200 mL and the final concn is 0.02mole/0.2L = ??
I hope this hasn't confused you. Check my work. It's getting late where I am.
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