Use the balanced equation to answer the following: 3AgNO3(aq) + FeCl3(aq)...Fe(NO3)3(aq) + 3AgCl(s) If 21.73g of FeCl3 is mixed with 48.97g of AgNO3

a) Which is the limiting reagent?
b) How many grams of AgCl can be theoretically produced?
c) If 15.97g of AgCl was collected, what is the percent yield for this reaction?

All of these stoichiometry problems are worked the same way although limiting reagent problems are slightly different (actually they are two stoichiometry problems rolled into one.)

148.6

To determine the limiting reagent, we need to calculate the amount of each reactant present and compare it to the stoichiometric ratio in the balanced equation.

Given:
Mass of FeCl3 = 21.73g
Mass of AgNO3 = 48.97g

a) To find the limiting reagent, we can calculate the number of moles of each reactant using their respective molar masses:

Molar mass of FeCl3 = 162.204 g/mol
Number of moles of FeCl3 = Mass of FeCl3 / Molar mass of FeCl3 = 21.73g / 162.204 g/mol = 0.134 moles

Molar mass of AgNO3 = 169.87 g/mol
Number of moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3 = 48.97g / 169.87 g/mol = 0.288 moles

Using the balanced equation, the stoichiometric ratio between FeCl3 and AgNO3 is 1:3. Therefore, for every 1 mole of FeCl3, 3 moles of AgNO3 are required.

Since the ratio of FeCl3 to AgNO3 is 1:3, and there are 0.134 moles of FeCl3 and 0.288 moles of AgNO3, we can see that AgNO3 is the limiting reagent. It is present in a lower amount compared to its stoichiometric requirement.

b) To determine the maximum amount of AgCl that can be produced, we need to use the stoichiometric ratio between AgCl and AgNO3 from the balanced equation. The ratio tells us that for every 3 moles of AgCl formed, 3 moles of AgNO3 are required.

Number of moles of AgNO3 = 0.288 moles
Using the stoichiometric ratio, the maximum moles of AgCl that can be produced is also 0.288 moles.

Molar mass of AgCl = 143.32 g/mol
Mass of AgCl = Number of moles of AgCl * Molar mass of AgCl = 0.288 moles * 143.32 g/mol = 41.26g

Therefore, the maximum amount of AgCl that can be theoretically produced is 41.26 grams.

c) To calculate the percent yield, we need to compare the actual yield (15.97g) to the theoretical yield (41.26g) and calculate the ratio:

Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (15.97g / 41.26g) * 100 = 38.66%

Therefore, the percent yield for this reaction is approximately 38.66%.

To determine the limiting reagent in this reaction, we need to compare the moles of each reactant and see which one is used up completely.

a) Calculate the moles of FeCl3 and AgNO3:

Molar mass of FeCl3 = 162.2 g/mol
Molar mass of AgNO3 = 169.9 g/mol

Moles of FeCl3 = 21.73 g FeCl3 / 162.2 g/mol FeCl3
Moles of AgNO3 = 48.97 g AgNO3 / 169.9 g/mol AgNO3

Calculating these values, we find:
Moles of FeCl3 = 0.134 mol
Moles of AgNO3 = 0.288 mol

From the balanced equation, we see that the stoichiometric ratio between FeCl3 and AgNO3 is 1:3. This means that 1 mole of FeCl3 reacts with 3 moles of AgNO3.

Since the stoichiometric ratio is 1:3, and we have fewer moles of FeCl3 (0.134 mol), FeCl3 is the limiting reagent.

b) To determine the maximum amount of AgCl that can be produced, we need to calculate the moles of AgCl produced using the limiting reagent. From the balanced equation, we know that 1 mole of FeCl3 produces 3 moles of AgCl.

So, moles of AgCl = moles of FeCl3 × 3 = 0.134 mol × 3 = 0.402 mol

Now, we can calculate the mass of AgCl using the molar mass of AgCl (143.3 g/mol):

Mass of AgCl = moles of AgCl × molar mass of AgCl
Mass of AgCl = 0.402 mol × 143.3 g/mol = 57.61 g

Therefore, the theoretical yield of AgCl is 57.61 grams.

c) To determine the percent yield of the reaction, we need to compare the actual yield (15.97 g) with the theoretical yield (57.61 g) and calculate the percentage:

Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (15.97 g / 57.61 g) × 100 = 27.75%

Therefore, the percent yield for this reaction is 27.75%.