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What is the anhydride for each of the following acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\), (b) \(\mathrm{HClO}_{3}\), (c) \(\mathrm{HNO}_{2}\), (d) \(\mathrm{H}_{2} \mathrm{CO}_{3}\), (e) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) ?

Short Answer

Expert verified
The anhydrides for the given acids are: (a) \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{7}\), (b) \(\mathrm{Cl}_{2}\mathrm{O}_{6}\), (c) \(\mathrm{N}_{2}\mathrm{O}_{3}\), (d) \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{5}\), and (e) \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}\).

Step by step solution

01

(a) Find the anhydride of \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

To find the anhydride for sulfuric acid, \(\mathrm{H}_{2}\mathrm{SO}_{4}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{H}_{2} \mathrm{SO}_{4} - \mathrm{H}_{2}\mathrm{O} = \mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{7}\) The anhydride for sulfuric acid is \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{7}\).
02

(b) Find the anhydride of \(\mathrm{HClO}_{3}\)

To find the anhydride for chloric acid, \(\mathrm{HClO}_{3}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{HClO}_{3} - \mathrm{H}_{2}\mathrm{O} = \mathrm{Cl}_{2}\mathrm{O}_{6}\) The anhydride for chloric acid is \(\mathrm{Cl}_{2}\mathrm{O}_{6}\).
03

(c) Find the anhydride of \(\mathrm{HNO}_{2}\)

To find the anhydride for nitrous acid, \(\mathrm{HNO}_{2}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{HNO}_{2} - \mathrm{H}_{2}\mathrm{O} = \mathrm{N}_{2}\mathrm{O}_{3}\) The anhydride for nitrous acid is \(\mathrm{N}_{2}\mathrm{O}_{3}\).
04

(d) Find the anhydride of \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

To find the anhydride for carbonic acid, \(\mathrm{H}_{2}\mathrm{CO}_{3}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{H}_{2}\mathrm{CO}_{3} - \mathrm{H}_{2}\mathrm{O} = \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{5}\) The anhydride for carbonic acid is \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{5}\).
05

(e) Find the anhydride of \(\mathrm{H}_{3} \mathrm{PO}_{4}\)

To find the anhydride for phosphoric acid, \(\mathrm{H}_{3}\mathrm{PO}_{4}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{H}_{3}\mathrm{PO}_{4} - \mathrm{H}_{2}\mathrm{O} = \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}\) The anhydride for phosphoric acid is \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}\).

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Key Concepts

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

Sulfuric Acid and Its Anhydride
Sulfuric acid, denoted as \(\mathrm{H}_2\mathrm{SO}_4\), is a highly important industrial chemical, known for its various uses in battery acid, fertilizer, and as a dehydrating agent. When it comes to its anhydride, the process involves combining two sulfuric acid molecules and removing a water molecule \(\mathrm{H}_2\mathrm{O}\), resulting in \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\), also known as pyrosulfuric acid or oleum. This is a useful reaction for understanding the relationship between acids and their respective anhydrides.

Oleum is used in various industrial applications, including the production of detergents and explosives. By understanding the creation of anhydrides, students can get insights into chemical synthesis and reactivity, both key concepts in chemistry.
Chloric Acid and Its Anhydride
Chloric acid, with the formula \(\mathrm{HClO}_3\), is known for its oxidative properties and is a part of a family of oxoacids of chlorine. To form its anhydride, you will combine two molecules of chloric acid and eliminate a single water molecule. This yields \(\mathrm{Cl}_2\mathrm{O}_6\), which is a powerful oxidizing agent itself.

Although less commonly encountered outside the laboratory, understanding chloric acid and its anhydride helps students grasp the varied oxidative states chlorine can exist in, offering a peek into the versatile nature of halogens in chemistry.
Nitrous Acid and Its Anhydride
Nitrous acid, \(\mathrm{HNO}_2\), is a weaker acid often used in the preparation of diazonium salts in organic synthesis. Its anhydride is formed by combining two nitrous acid molecules and removing one molecule of water, resulting in \(\mathrm{N}_2\mathrm{O}_3\), known as dinitrogen trioxide. This compound is important as it exemplifies the behavior of nitrogen oxides, which are significant in the study of environmental chemistry and atmospheric reactions.

It's critical to understand the properties and reactions of nitrous acid and its anhydride as they play a significant role in nitrogen cycles and pollution.
Carbonic Acid and Its Anhydride
Carbonic acid \(\mathrm{H}_2\mathrm{CO}_3\) plays an essential role in biological and geological processes. Like others, its anhydride is constructed by removing a water molecule from two carbonic acid molecules, which gives \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_5\), though it's more commonly recognized in the form of its gaseous counterpart, carbon dioxide \(\mathrm{CO}_2\).

This transformation is pivotal in understanding the carbon cycle and the impact of \(\mathrm{CO}_2\) in environmental science, as well as the role of acids and anhydrides in biochemical systems.
Phosphoric Acid and Its Anhydride
Phosphoric acid \(\mathrm{H}_3\mathrm{PO}_4\) is widely used in agriculture and food industries. Its anhydride, \(\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7\), or pyrophosphoric acid, results from dehydrating two phosphoric acid molecules. This occurs naturally when phosphoric acid is heated, leading to the loss of a water molecule.

The anhydride form is significant in biochemistry, especially in DNA synthesis and energy transfer through ATP. For students, understanding the process of forming pyrophosphoric acid can help illustrate the concept of condensation reactions and their importance in both industrial chemistry and life sciences.

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Most popular questions from this chapter

SOLUTION Analyze We must write balanced nuclear equations in which the masses and charges of reactants and products are equal. Plan We can begin by writing the complete chemical symbols for the nuclei and decay particles that are given in the problem. Solve (a) The information given in the question can be summarized as $$ { }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{Z}^{A} \mathrm{X} $$ The mass numbers must have the same sum on both sides of the equation: $$ 201+0=A $$ Thus, the product nucleus must have a mass number of 201. Similarly, balancing the atomic numbers gives $$ 80-1=Z $$ Thus, the atomic number of the product nucleus must be 79 , which identifies it as gold (Au): $$ { }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{79}^{201} \mathrm{Au} $$ (b) In this case we must determine what type of particle is emitted in the course of the radioactive decay: $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{\mathrm{Z}}^{\mathrm{X}} \mathrm{X} $$ From \(231=231+A\) and \(90=91+Z\), we deduce \(A=0\) and \(Z=-1\). According to Table \(21.2\), the particle with these characteristics is the beta particle (electron). We therefore write $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{-1}^{0} \mathrm{e} \text { or } \quad{ }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+\beta^{-} $$

\text { Write the balanced nuclear equation for the process summarized as }{ }_{13}^{27} \mathrm{Al}(\mathrm{n}, \alpha)_{11}^{24} \mathrm{Na} \text {. }SOLUTION Analyze We must go from the condensed descriptive form of the reaction to the balanced nuclear equation. Plan We arrive at the balanced equation by writing \(\mathrm{n}\) and \(\alpha\), each with its associated subscripts and superscripts. Solve The \(\mathrm{n}\) is the abbreviation for a neutron \(\left({ }_{0}^{1} \mathrm{n}\right)\) and \(\alpha\) represents an alpha particle ( \(\left.{ }_{2}^{4} \mathrm{He}\right)\). The neutron is the bombarding particle, and the alpha particle is a product. Therefore, the nuclear equation is $$ { }_{13}^{27} \mathrm{Al}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{11}^{24} \mathrm{Na}+{ }_{2}^{4} \mathrm{He} \text { or } \quad{ }_{13}^{27} \mathrm{Al}+\mathrm{n} \longrightarrow{ }_{11}^{24} \mathrm{Na}+\alpha $$

a) Determine the number of sodium ions in the chemical formula of albite, \(\mathrm{Na}_{x} \mathrm{AlSi}_{3} \mathrm{O}_{g}\). (b) Determine the number of hydroxide ions in the chemical formula of tremolite, \(\mathrm{Ca}_{2} \mathrm{Mg}_{5}\left(\mathrm{Si}_{4} \mathrm{O}_{11}\right)_{2}(\mathrm{OH})_{\mathrm{x}}\) -

Give the chemical formula for (a) hydrocyanic acid, (b) nickel tetracarbonyl, (c) barium bicarbonate, (d) calcium acetylide, (e) potassium carbonate.

The maximum allowable concentration of \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) in air is \(20 \mathrm{mg}\) per kilogram of air (20 ppm by mass). How many grams of FeS would be required to react with hydrochloric acid to produce this concentration at \(1.00\) atm and \(25^{\circ} \mathrm{C}\) in an average room measuring \(12 \mathrm{ft} \times 20 \mathrm{ft} \times 8 \mathrm{ft}\) ? (Under these conditions, the average molar mass of air is \(29.0 \mathrm{~g} / \mathrm{mol}\).)

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