The class will react 50.00 mL samples of .200 mol/L potassium phosphate with an excess of 0.120 mol/L lead II nitrate solution.

what is the minimum volume of lead II nitrate solution required? What volume of lead II nitrate solution should the instructor tell the students to use? Describe how the students can test for completeness of reaction of the limiting reagent?

I am really stuck any help would be greatly appreciated! Thanks.

To answer these questions, we need to use stoichiometry and the concept of limiting reactants.

1. Minimum volume of lead II nitrate solution required:
To find the minimum volume of lead II nitrate solution required, we need to determine which reactant is the limiting reagent. In other words, which reactant will be completely consumed in the reaction.

First, we need to write the balanced chemical equation for the reaction between potassium phosphate (K3PO4) and lead II nitrate (Pb(NO3)2):

3K3PO4 + 2Pb(NO3)2 → 6KNO3 + Pb3(PO4)2

From the balanced equation, we can see that the molar ratio between potassium phosphate and lead II nitrate is 3:2. This means that for every 3 moles of K3PO4, we need 2 moles of Pb(NO3)2.

Now, let's calculate the moles of each reactant in the given sample:

Moles of K3PO4 = concentration (mol/L) × volume (L)
= 0.200 mol/L × 0.0500 L
= 0.010 mol

Since the molar ratio between K3PO4 and Pb(NO3)2 is 3:2, we can calculate the moles of Pb(NO3)2 required:

Moles of Pb(NO3)2 required = (2 moles Pb(NO3)2 / 3 moles K3PO4) × 0.010 mol K3PO4
= 0.00667 mol

Next, calculate the volume of Pb(NO3)2 solution required:

Volume (L) = moles / concentration
= 0.00667 mol / 0.120 mol/L
≈ 0.056 L
≈ 56.0 mL

Therefore, the minimum volume of lead II nitrate solution required is approximately 56.0 mL.

2. Volume of lead II nitrate solution the instructor should tell the students to use:
Since the minimum volume required is approximately 56.0 mL, the instructor should tell the students to use a volume slightly greater than that, to ensure that there is an excess of lead II nitrate solution present. For example, the instructor could tell the students to use 60.0 mL or 65.0 mL of lead II nitrate solution.

3. Testing for completeness of the reaction:
To test for the completeness of the reaction, the students can add a few drops of a solution that would react with any remaining lead II nitrate. In this case, they can add a few drops of a solution containing chloride ions (e.g., hydrochloric acid or sodium chloride solution).

If any lead II nitrate remains unreacted, it will react with the chloride ions to form a white precipitate of lead II chloride (PbCl2). If the reaction is complete, no further precipitation will occur.

The students can also confirm the completeness of the reaction by checking whether the expected amount of products has been formed, according to the balanced equation. In this case, the students can calculate the theoretical yield of the product lead II phosphate (Pb3(PO4)2) based on the moles of the limiting reactant (potassium phosphate) used. Then, they can compare the actual amount of lead II phosphate obtained in grams to the theoretical yield. If the actual amount is close to the theoretical yield, it indicates that the reaction is complete.

Remember, it's important for students to take safety precautions and consult their instructor before conducting any experiments in the laboratory.

To determine the minimum volume of lead II nitrate solution required, we need to use the stoichiometry of the balanced chemical equation between potassium phosphate (K3PO4) and lead II nitrate (Pb(NO3)2).

The balanced chemical equation for the reaction between K3PO4 and Pb(NO3)2 is:

3K3PO4 + 2Pb(NO3)2 -> 6KNO3 + Pb3(PO4)2

From the equation, we can see that the mole ratio between K3PO4 and Pb(NO3)2 is 3:2. This means that for every 3 moles of K3PO4, we need 2 moles of Pb(NO3)2.

Given that the reaction is being carried out using 0.200 mol/L K3PO4 and an excess of 0.120 mol/L Pb(NO3)2, we can use this information to determine the minimum volume of Pb(NO3)2 solution required.

Step 1: Calculate the moles of K3PO4:
Moles of K3PO4 = concentration (mol/L) x volume (L)
Moles of K3PO4 = 0.200 mol/L x 0.050 L = 0.010 mol

Step 2: Determine the moles of Pb(NO3)2 required:
Moles of Pb(NO3)2 = (moles of K3PO4) x (2 moles of Pb(NO3)2 / 3 moles of K3PO4)
Moles of Pb(NO3)2 = 0.010 mol x (2/3) = 0.0067 mol

Step 3: Calculate the volume of Pb(NO3)2 solution required:
Volume of Pb(NO3)2 solution = (moles of Pb(NO3)2) / (concentration of Pb(NO3)2)

For the minimum volume:
Volume of Pb(NO3)2 solution = 0.0067 mol / 0.120 mol/L = 0.056 L = 56 mL

Therefore, the minimum volume of Pb(NO3)2 solution required is 56 mL.

To ensure a complete reaction, the instructor should instruct the students to use a slightly excess volume of Pb(NO3)2 solution. This will guarantee that all the K3PO4 is reacted and prevent any unreacted K3PO4 from being left over.

For testing completeness of reaction, students can perform a simple precipitation test. They can add a few drops of a solution containing an anion that forms an insoluble salt with any excess Pb2+ ions (e.g., sodium sulfate or sodium chloride). If a precipitate forms, it indicates that the reaction is not complete, and more Pb(NO3)2 solution is needed. If no precipitate forms, it suggests that the reaction is complete.