| Reactions in Aqueous Solution | |
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Problem
Mixing 20.0 cm3 of 0.100 M AgNO3 with 15.0 cm3 of 0.200 M NaCl produces the following reaction:
Ag+(aq) + Cl-(aq) --> AgCl(s)
a. Determine the limiting reactant.
Solution
Begin by calculating the numbers of moles of Ag+ and Cl- that are present in the solutions before mixing. Use that information to calculate the actual ratio of moles Ag+ to moles Cl-. This ratio is compared to the stoichiometric ratio in order to determine the limiting reactant. Then, the theoretical yield of AgCl will be based on the amount of limiting reactant.
a. One mole of AgNO3 yields one mole of Ag+, so the number of moles of AgNO3 equals the number of moles of Ag+. Similarly, the number of moles of Cl- equals the number of moles of NaCl. Therefore:
moles Ag+ = 0.200 liters x 0.100 mol/liter
moles Ag+ = 2.00 x 10-3 mol Ag+
moles Cl- = 0.0150 liters x 0.200 mol/liter
moles Cl- = 3.00 x 10-3 mol Cl-
The actual ratio is 2.00 x 10-3 mol Ag+ / 3.00 x 10-3 mol Cl- = 0.667 mol Ag+/mol Cl-
The stoichiometric ratio is 1 mol Ag+ /1 mol Cl-. Since the actual ratio is less than the stoichiometric ratio, there insufficient Ag+ to react with all of the Cl-, so the limiting reactant is AgNO3.
b. Theoretical yield of AgCl is based on the amount of Ag+ that is present. Using the Periodic Table, we determine that one mole of AgCl weighs 143.3 g. Plugging in these numbers gives us:
theoretical yield AgCl = 2.00 x 10-3 mol Ag+ x 1 mol AgCl/1 mol Ag+ x 143.3 g AgCl/1 mol AgCl
theoretical yield AgCl = 0.287 g AgCl
Answer
a. AgNO3 Previous page > Sample Problem #1 > Page 1, 2
b. Determine the theoretical yield of AgCl in grams.
b.0.287 g AgCl
