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Chapter 22: Problem 81

The lead(IV) oxide used in car batteries is prepared by coating an electrode plate with \(\mathrm{PbO}\) and then oxidizing it to lead dioxide \(\left(P b O_{2}\right)\). Despite its name, \(P b O_{2}\) has a nonstoichiometric mole ratio of lead to oxygen of about \(1 / 1.98 .\) In fact, the holes in the \(\mathrm{PbO}_{2}\) crystal structure due to missing \(\mathrm{O}\) atoms are responsible for the oxide's conductivity. (a) What is the mole \(\%\) of \(\mathrm{O}\) missing from the \(\mathrm{PbO}_{2}\) structure? (b) What is the molar mass of the nonstoichiometric compound?

Short Answer

Expert verified
The mole percentage of missing O is 1%. The molar mass is 238.48 g/mol.

Step by step solution

01

Understand the problem

First, grasp the task: Find the mole percentage of O missing and the molar mass of the nonstoichiometric compound.
02

Calculate the Expected Number of Oxygen Atoms

For the stoichiometric compound \( \mathrm{PbO}_2 \), the expected ratio is 1 mole of Pb to 2 moles of O.
03

Use the Given Ratio

The given ratio of Pb to O is 1:1.98. That means there's a slight deficiency of O atoms compared to the expected 2 in pure \(\mathrm{PbO}_2\).
04

Calculate the Deficiency in Oxygen

The deficiency in O atoms is \(2 - 1.98 = 0.02 \) moles of O.
05

Determine the Mole Percentage of Missing Oxygen

The mole percentage of missing oxygen is \(\left( \frac{0.02}{2} \right) \times 100 = 1 \% \).
06

Calculate the Atomic Weights

The atomic masses are: Pb = 207.2 g/mol and O = 16.00 g/mol.
07

Determine the Molar Mass of the Nonstoichiometric Compound

Using the ratio Pb to O (1:1.98), calculate the molar mass: \( 207.2 + 1.98 \times 16 = 238.48 \) g/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lead Dioxide
Lead dioxide, denoted as \(\mathrm{PbO}_2\), is a chemical compound of lead in its +4 oxidation state. This compound is commonly used in the electrochemical cells of car batteries. It is prepared by first coating an electrode plate with lead oxide \(\mathrm{PbO}\) and then oxidizing it to form lead dioxide. Although the ideal stoichiometric ratio for lead dioxide suggests a 1:2 ratio of lead to oxygen atoms, in reality, it often shows a nonstoichiometric nature with a slightly deficient amount of oxygen atoms due to crystal structure quirks, contributing to its conductivity properties.
Molar Mass Calculation
Calculating the molar mass involves summing up the atomic masses of all atoms in a compound. For lead dioxide, assuming an ideal stoichiometry, the molar mass calculation is straightforward using the atomic masses: Lead (Pb) is 207.2 g/mol, and Oxygen (O) is 16.00 g/mol. For nonstoichiometric compounds like \(\mathrm{PbO}_2\) with a ratio of 1:1.98:
$$207.2 + 1.98 \times 16 = 238.48$$ g/mol. This calculation helps chemists understand the total mass one mole of the compound would have, which is crucial for stoichiometric calculations in reactions.
Oxidation Reaction
An oxidation reaction involves the loss of electrons by a molecule, atom, or ion. In the case of lead dioxide production, \(\mathrm{PbO}\) is oxidized to \(\mathrm{PbO}_2\). Here, the lead in \(\mathrm{PbO}\) (where it is in a +2 oxidation state) is transformed into lead dioxide \(\mathrm{PbO}_2\) (where lead has a +4 oxidation state). This process can be summarized with the equation:
$$ Pb^{2+} + O_2 \rightarrow Pb^{4+}O_2 $$. Understanding this reaction is essential as it underpins the functionality of lead-acid batteries by providing a means to store and discharge electrical energy.
Deficiency in Oxygen
Deficiency in oxygen in lead dioxide usually refers to the situation where the actual amount of oxygen in the compound is less than what is expected from its ideal stoichiometric formula. For \(\mathrm{PbO}_2\), the nonstoichiometric ratio is about 1:1.98 instead of 1:2. This small deficiency, calculated as \(2 - 1.98 = 0.02\) moles of O, implies 1% of the oxygen is missing. This minute change hugely impacts the structural and conductive properties of lead dioxide. The presence of oxygen vacancies can impart increased electrical conductivity, making such nonstoichiometric compounds very valuable for specific industrial and technological applications.

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Most popular questions from this chapter

The overall cell reaction for aluminum production is $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}\left(\text { in } \mathrm{Na}_{3} \mathrm{AlF}_{6}\right)+3 \mathrm{C}(\mathrm{graphite}) \longrightarrow 4 \mathrm{Al}(t)+3 \mathrm{CO}_{2}(g) $$ (a) Assuming \(100 \%\) efficiency, how many metric tons (t) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) are consumed per metric ton of Al produced? (b) Assuming \(100 \%\) efficiency, how many metric tons of the graphite anode are consumed per metric ton of Al produced? (c) Actual conditions in an aluminum plant require \(1.89 \mathrm{t}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(0.45 \mathrm{t}\) of graphite per metric ton of Al. What is the percent yicld of \(\mathrm{Al}\) with respect to \(\mathrm{Al}_{2} \mathrm{O}_{3} ?\) (d) What is the percent yield of Al with respect to graphite? (e) What volume of \(\mathrm{CO}_{2}\) (in \(\mathrm{m}^{3}\) ) is produced per metric ton of Al at operating conditions of \(960 .{ }^{\circ} \mathrm{C}\) and exactly 1 atm?

The final step in the smelting of CuFeS \(_{2}\) is $$ \mathrm{Cu}_{2} \mathrm{~S}(s)+2 \mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow 6 \mathrm{Cu}(I)+\mathrm{SO}_{2}(g) $$ (a) Give the oxidation states of copper in \(\mathrm{Cu}_{2} \mathrm{~S}, \mathrm{Cu}_{2} \mathrm{O},\) and \(\mathrm{Cu} .\) (b) What are the oxidizing and reducing agents in this reaction?

Even though most metal sulfides are sparingly soluble in water, their solubilities differ by several orders of magnitude. This difference is sometimes used to separate the metals by controlling the pH. Use the following data to find the pH at which you can separate \(0.10 M \mathrm{Cu}^{2+}\) and \(0.10 \mathrm{M} \mathrm{Ni}^{2+}\) Saturated \(\mathrm{H}_{2} \mathrm{~S}=0.10 \mathrm{M}\) \(K_{a t}\) of \(\mathrm{H}_{2} \mathrm{~S}=9 \times 10^{-8} \quad K_{a 2}\) of \(\mathrm{H}_{2} \mathrm{~S}=1 \times 10^{-17}\) \(K_{v}\) of NiS \(=1.1 \times 10^{-18} \quad K_{\mathrm{sp}}\) of \(\mathrm{CuS}=8 \times 10^{-34}\)

22.66 Why isn't nitric acid produced by oxidizing \(\mathrm{N}_{2}\) as follows? (1) $$ \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ (2) \(3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\) (3) \(\frac{2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)}{3 \mathrm{~N}_{2}(g)+6 \mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{HNO}_{3}(a q)+2 \mathrm{NO}(g)}\) (Hint: Evaluate the thermodynamics of cach step.)

World production of chromite \(\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right),\) the main ore of chromium, was \(1.97 \times 10^{7}\) metric tons in 2006 . To isolate chromium, a mixture of chromite and sodium carbonate is heated in air to form sodium chromate, iron(III) oxide, and carbon dioxide. The sodium chromate is dissolved in water, and this solution is acidified with sulfuric acid to produce the less soluble sodium dichromate. The sodium dichromate is filtered out and reduced with carbon to produce chromium(III) oxide, sodium carbonate, and carbon monoxide. The chromium(III) oxide is then reduced to chromium with aluminum metal. (a) Write balanced equations for each step. (b) What mass of chromium (in \(\mathrm{kg}\) ) could be prepared from the 2006 world production of chromite?
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