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Uranium hexafluoride, \(\mathrm{UF}_{6},\) is processed to produce fuel for nuclear reactors and nuclear weapons. UF \(_{6}\) can be produced in a two-step reaction. Solid uranium (IV) oxide, \(\mathrm{UO}_{2}\), is first made to react with hydrofluoric acid (HF) solution to form solid UF \(_{4}\) with water as a by-product. UF \(_{4}\) further reacts with fluorine gas to form \(\mathrm{UF}_{6}\). (a) Write the balanced molecular equations for the conversion of \(\mathrm{UO}_{2}\) into \(\mathrm{UF}_{4}\) and the conversion of \(\mathrm{UF}_{4}\) to \(\mathrm{UF}_{6}\). (b) Which step is an acid-base reaction? (c) Which step is a redox reaction?

Short Answer

Expert verified
(a) 1. \(\mathrm{UO}_{2} + 4\, \mathrm{HF} \rightarrow \mathrm{UF}_{4} + 2\, \mathrm{H}_{2}\mathrm{O}\); 2. \(\mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6}\). (b) First step. (c) Second step.

Step by step solution

01

Write Balanced Equation for First Reaction

The first reaction involves uranium (IV) oxide, \(\mathrm{UO}_{2}\), reacting with hydrofluoric acid (HF) to form uranium tetrafluoride (\(\mathrm{UF}_{4}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). The balanced chemical equation is: \[\mathrm{UO}_{2}(s) + 4\, \mathrm{HF}(aq) \rightarrow \mathrm{UF}_{4}(s) + 2\, \mathrm{H}_{2}\mathrm{O}(l)\].
02

Write Balanced Equation for Second Reaction

The second reaction involves \(\mathrm{UF}_{4}\) reacting with fluorine gas (F\(_2\)) to form uranium hexafluoride (\(\mathrm{UF}_{6}\)). The balanced chemical equation is: \[\mathrm{UF}_{4}(s) + \mathrm{F}_{2}(g) \rightarrow \mathrm{UF}_{6}(g)\].
03

Identify Acid-Base Reaction

An acid-base reaction is characterized by the transfer of protons (H\(^+\)). In the first reaction, \(\mathrm{HF}\), an acid, donates protons to form \(\mathrm{H}_{2}\mathrm{O}\). Hence, the first step (\(\mathrm{UO}_{2}\) with \(\mathrm{HF}\)) is the acid-base reaction.
04

Identify Redox Reaction

A redox reaction involves the transfer of electrons. The second reaction involves \(\mathrm{F}_{2}\), a strong oxidizing agent, reacting with \(\mathrm{UF}_{4}\) to add more fluoride ions and change the oxidation state of uranium. Hence, the second step (\(\mathrm{UF}_{4}\) with \(\mathrm{F}_{2}\)) is the redox reaction.

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Key Concepts

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

Understanding Acid-Base Reactions
In chemistry, acid-base reactions are pivotal as they involve the exchange of protons (H\(^+\)) between reactants. This is a hallmark of Bronsted-Lowry theory. When looking at the conversion of uranium (IV) oxide (UO\(_2\)) with hydrofluoric acid (HF) to produce uranium tetrafluoride (UF\(_4\)) and water, we see a clear acid-base reaction. In this process:
  • Hydrofluoric acid (HF) acts as the acid as it donates protons (H\(^+\)) to form water (H\(_2\)O).
  • The uranium oxide (UO\(_2\)) effectively serves as the base by accepting these protons. However, rather than bonding directly with protons, it facilitates the formation of UF\(_4\).
  • This exchange results in the formation of water, a classic by-product of acid-base reactions.
Recognizing these characteristics can help you identify an acid-base reaction among other chemical processes. This understanding is foundational in chemistry and is widely applicable beyond this specific context.
As you delve into more complex reactions, keeping track of proton transfer will guide you in determining the roles of acids and bases.
The Basics of Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are essential as they involve the transfer of electrons between substances. This movement of electrons leads to changes in oxidation states, which are key to identifying these reactions. In the reaction where uranium tetrafluoride (UF\(_4\)) is converted to uranium hexafluoride (UF\(_6\)) using fluorine gas (F\(_2\)), we observe a redox reaction. Here’s how it works:
  • Oxidation: The fluorine gas accepts electrons, causing the oxidation number of the uranium in UF\(_4\) to increase, as more fluoride ions are added to form UF\(_6\).
  • Reduction: As fluorine gains electrons, its oxidation number decreases. This dual process—one substance losing electrons while another gains—defines a redox reaction.
  • Redox reactions are critical in energy transfer and many industrial processes, making them a fundamental aspect of chemical reactions and technology.
Understanding these processes clarifies the transitions that elements undergo as they gain or lose electrons. It highlights the role of electron transfer in chemical changes and helps predict product formation in reactions where redox is involved.
Balancing Chemical Equations
A balanced chemical equation is crucial in chemistry as it reflects the conservation of mass, where the number of atoms of each element is the same on both sides of the equation. For the reactions involving uranium compounds:
  • In the first reaction, UO\(_2\) reacts with 4 moles of HF to produce UF\(_4\) and 2 moles of water (H\(_2\)O), illustrating that every reactant's atom count is conserved in the products.
  • In the second reaction, adding F\(_2\) to UF\(_4\) yields UF\(_6\), showing again how the fluorine added must balance out with no leftover atoms.
  • This balance ensures that equations accurately represent the quantifiable relationships involved, thereby allowing chemists to predict how substances will react and in what proportions.
Balancing equations requires practice. Start from one element at a time and adjust coefficients to ensure both sides of the equation match. This skill not only aids in understanding chemical reaction dynamics but is also essential in laboratories, industries, and any chemical-related field.

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Most popular questions from this chapter

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}^{3-},\) what mass of sodium arsenate would be present in a \(1.00-\mathrm{L}\) sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9} $$

Acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3},\) is a nonelectrolyte; hypochlorous acid, HClO, is a weak electrolyte; and ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a strong electrolyte. (a) What are the solutes present in aqueous solutions of each compound? (b) If 0.1 mol of each compound is dissolved in solution, which one contains 0.2 mol of solute particles, which contains 0.1 mol of solute particles, and which contains somewhere between 0.1 and 0.2 mol of solute particles?

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, HCOOH.

A sample of \(8.69 \mathrm{~g}\) of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(155.0 \mathrm{~mL}\) of \(0.750 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\). (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Zn}(\mathrm{OH})_{2}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{ZnSO}_{4}\) are present after the reaction is complete?

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for 10.0 min at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) \((\mathbf{c})\) Is this manufacturer in compliance with regulations?

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