# Chemistry

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A student mixes 100.0 mL of 0.300 mol/L silver nitrate solution with 100.0 mL of 0.300 mol/L calcium chloride solution.

a) Determine the theoretical yield of the precipitate.

b) If the actual mass of precipitate is 4.00g, calculate the percent error of this experiment.

• Chemistry - ,

Here is a solved example stoichiometry problem which provides the theoretical yield. Percent error, in this case, is the same as 100%-%yield. Post your work if you get stuck.
http://www.jiskha.com/science/chemistry/stoichiometry.html

• Chemistry - ,

this is what i have so far:

2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2

for AgNO3:

V= 0.100L
c=0.0300 mol/L
n=cV
= 0.03 mol

for CaCl2:

V= 0.100L
c=0.0300 mol/L
n=cV
= 0.03 mol

and now I don't know how to determine the limiting reactant.

• Chemistry - ,

You have made two correcting errors so far.
M AgNO3 in the problem is 0.300 M (not 0.03) but the second error corrects that.
0.1 x 0.300 = 0.0300 moles AgNO3 which is correct
The same two errors occur for CaCl2 but you have arrived at the correct answer of 0.0300 moles CaCl2.
You just do two simple stoichiometry problems when you have a limiting reagent.
2AgNO3 + CaCl2 ==> Ca(NO3)2 + 2AgCl.

Convert moles AgNO3 to moles AgCl
0.0300 moles AgNO3 x (2 moles AgCl/2 moles AgNO3) = 0.03 moles AgCl produced.
AND
Convert moles CaCl2 t moles AgCl.
0.0300 moles CaCl2 x (2 moles AgCl/1 mole CaCl2) = 0.0300*2 = 0.0600 moles AgCl produced.
You note TWO answers for amount AgCl produced; obviously one of them must be wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing this value is the limiting reagent.