   Chapter 20, Problem 99CP

Chapter
Section
Textbook Problem

# Lead forms compounds in the +2 and +4 oxidation states. All lead(II) halides are known (and are known to be ionic). Only PbF4 and PbCl4 are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: PbX 4 → PbX 2 + X 2 Suppose 25.00 g of a lead(IV) halide reacts to form 16.12 g of a lead(II) halide and the free halogen. Identify the halogen.

Interpretation Introduction

Interpretation: The decomposition reaction of lead (IV) halide (PbX4) , mass of PbX4 and PbX2 is given. The halogen X is to be identified.

Concept introduction: The number of moles is calculated using the formula,

Numberofmoles=GivenmassMolar mass of the compound

Molar mass is the sum of each element of a compound multiplied by the number of atoms present.

To determine: The halogen X in the decomposition reaction of lead (IV) halide.

Explanation

Explanation

Given

Mass of PbX4 is 25.00g .

Mass of PbX2 is 16.12g .

The stated reaction is,

PbX4PbX2+X2

From the above equation, it is clear that, number of moles of PbX2 is equal to the number of moles of PbX4 .

It is assumed that, the molar mass of X is M and the number of moles of PbX4 is n .

The molar mass of PbX4 is,

207.2g/mol+4×M

The molar mass of PbX2 is,

207.2g/mol+2×M

The number of moles is calculated using the formula,

Numberofmoles=GivenmassMolar mass of the compound (1)

Substitute the values of mass, number of moles and molar mass of PbX4 in the above equation.

Numberofmoles=GivenmassMolar mass of the compoundn=25.0g207.2g/mol+4×Mn(207

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