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.
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
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.
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.
To determine the theoretical yield of the precipitate, we need to find out which compound will precipitate when silver nitrate and calcium chloride are mixed, and then calculate the amount of the precipitate formed using stoichiometry.
a) To find out the compound that will precipitate, we need to identify the possible products of the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2). In this case, the insoluble compound formed is silver chloride (AgCl). The balanced equation for the reaction can be written as:
AgNO3 + CaCl2 â AgCl + Ca(NO3)2
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of CaCl2 to produce 1 mole of AgCl.
Given that the initial volume for both solutions is 100.0 mL and the molarity (concentration) of each solution is 0.300 mol/L, we can calculate the number of moles of AgNO3 and CaCl2:
moles of AgNO3 = volume (L) x concentration (mol/L) = 0.100 L x 0.300 mol/L = 0.0300 mol
moles of CaCl2 = volume (L) x concentration (mol/L) = 0.100 L x 0.300 mol/L = 0.0300 mol
Since the stoichiometry of the reaction is 1:1, the limiting reagent is AgNO3 because it produces an equal number of moles of AgCl. Therefore, the theoretical yield of AgCl is 0.0300 mol.
To convert the moles of AgCl to grams, we need to know the molar mass of AgCl, which can be calculated as follows:
molar mass of AgCl = atomic mass of Ag + atomic mass of Cl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Now we can calculate the theoretical yield in grams:
theoretical yield = moles of AgCl x molar mass of AgCl
theoretical yield = 0.0300 mol x 143.32 g/mol = 4.2996 g
Therefore, the theoretical yield of the precipitate (AgCl) is approximately 4.30 grams.
b) To calculate the percent error, we need to compare the actual mass of the precipitate to the theoretical yield.
percent error = |(actual yield - theoretical yield) / theoretical yield| x 100
Given that the actual mass of the precipitate is 4.00 grams and the theoretical yield is 4.30 grams, we can calculate the percent error:
percent error = |(4.00 g - 4.30 g) / 4.30 g| x 100 = |-0.30 g / 4.30 g| x 100 = 6.98%
Therefore, the percent error of this experiment is approximately 6.98%.