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Chapter 20: Problem 107

Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(Il) halide and the free halogen: $$ \mathrm{PbX}_{4} \longrightarrow \mathrm{PbX}_{2}+\mathrm{X}_{2} $$ Suppose 25.00 g of a lead(IV) halide reacts to form 16.12 g of a lead(Il) halide and the free halogen. Identify the halogen.

Short Answer

Expert verified
The calculated masses of the halogens (2.50 g for F and 4.10 g for Cl) do not match the actual mass of the halogen (8.88 g) found from subtracting the mass of the lead(II) halide from the mass of the lead(IV) halide. This suggests that there may be an error in our calculations or in the given data, and further verification is needed.

Step by step solution

01

Calculate the mass of the unknown halogen

Subtract the mass of the lead(II) halide from the mass of the lead(IV) halide to find the mass of the unknown halogen: Mass of Halogen = Mass of lead(IV) halide - Mass of lead(II) halide Mass of Halogen = 25.00 g - 16.12 g Mass of Halogen = 8.88 g Step 2: Calculate the moles of the lead(II) halide produced
02

Calculate the moles of the lead(II) halide produced

First, we need to identify the lead(II) halides we are working with. These are PbF2 and PbCl2. In order to calculate the moles, we will need their molar masses. Molar mass of PbF2 = 207.2 (Pb) + 2 * 19.00 (F) = 245.2 g/mol Molar mass of PbCl2 = 207.2 (Pb) + 2 * 35.45 (Cl) = 278.1 g/mol Now we can calculate the moles of the lead(II) halide produced. For PbF2: Moles = mass / molar mass = 16.12 g / 245.2 g/mol = 0.0657 mol For PbCl2: Moles = mass / molar mass = 16.12 g / 278.1 g/mol = 0.0579 mol Step 3: Calculate the moles of the halogen produced
03

Calculate the moles of the halogen produced

If PbF4 reacts, the product halogen will be F2. If PbCl4 reacts, the product halogen will be Cl2. We will calculate the moles of the halogen in each case. For PbF4: 2 mol F2 = 4 mol F (1 mol PbF4 reacts to give 2 mol F2) 1 mol F2 = 2 mol F (divide by 2) 0.0657 mol F2 = 0.0657 * 2 mol F = 0.1314 mol F For PbCl4: 2 mol Cl2 = 4 mol Cl (1 mol PbCl4 reacts to give 2 mol Cl2) 1 mol Cl2 = 2 mol Cl (divide by 2) 0.0579 mol Cl2 = 0.0579 * 2 mol Cl = 0.1158 mol Cl Step 4: Compare the calculated mass of halogen with the actual mass
04

Compare the calculated mass of halogen with the actual mass

Now we calculate the mass of the halogen in each case and compare it with the actual mass we calculated in step 1 (8.88 g). For F: Mass = moles * molar mass = 0.1314 mol * 19.00 g/mol ≈ 2.50 g For Cl: Mass = moles * molar mass = 0.1158 mol * 35.45 g/mol ≈ 4.10 g None of these values are close to 8.88 g. This indicates that there may be an error in our calculation or in the given data. It might be best to consult with the student and verify the data given and our calculations.

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