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Chapter 22: Problem 39

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(s)\), reacts with ozone to form \(\mathrm{PbSO}_{4}(s)\) and \(\mathrm{O}_{2}(g)\). (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to \(\mathrm{ZnO}\). (e) Potassium peroxide reacts (f) Oxygen with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\). is converted to ozone in the upper atmosphere.

Short Answer

Expert verified
(a) \(2\mathrm{HgO} \rightarrow \mathrm{O}_{2} + 2\mathrm{Hg}\) (b) \(\mathrm{Cu(NO_3)_2} \rightarrow \mathrm{CuO} + 2\mathrm{NO_2} + \frac{1}{2}\mathrm{O_2}\) (c) \(\mathrm{PbS} + 2\mathrm{O_3} \rightarrow \mathrm{PbSO_4} + 2\mathrm{O_2}\) (d) \(\mathrm{ZnS} + \frac{1}{2}\mathrm{O_2} \rightarrow \mathrm{ZnO}\) (e) \(\mathrm{K_2O_2} + \mathrm{CO_2} \rightarrow \mathrm{K_2CO_3} + \mathrm{O_2}\) (f) \(\frac{3}{2}\mathrm{O_2} \rightarrow \mathrm{O_3}\)

Step by step solution

01

a) Decomposition of mercury(II) oxide

To write the balanced equation for the decomposition of mercury(II) oxide, first write the formula for the reactant (mercury(II) oxide) and the products (oxygen and mercury metal) and then find the coefficients that balance the atoms. The formula for mercury(II) oxide is \(\mathrm{HgO}\), and mercury metal is represented by \(\mathrm{Hg}\). The decomposition reaction can be written as: $$\mathrm{HgO} \rightarrow \mathrm{O}_{2} + \mathrm{Hg}$$ To balance the equation, we need 2 \(\mathrm{HgO}\) molecules to produce 2 \(\mathrm{Hg}\) atoms and 1 \(\mathrm{O}_{2}\) molecule: $$2\mathrm{HgO} \rightarrow \mathrm{O}_{2} + 2\mathrm{Hg}$$
02

(b) Decomposition of copper(II) nitrate

Copper(II) nitrate has the formula \(\mathrm{Cu(NO_3)_2}\) and decomposes into copper(II) oxide (\(\mathrm{CuO}\)), nitrogen dioxide (\(\mathrm{NO_2}\)), and oxygen (\(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{Cu(NO_3)_2} \rightarrow \mathrm{CuO} + \mathrm{NO_2} + \mathrm{O_2}$$ To balance the equation, we need 2 \(\mathrm{NO_2}\) molecules and 1 \(\mathrm{O_2}\) molecule for each \(\mathrm{Cu(NO_3)_2}\) molecule decomposed: $$\mathrm{Cu(NO_3)_2} \rightarrow \mathrm{CuO} + 2\mathrm{NO_2} + \frac{1}{2}\mathrm{O_2}$$
03

(c) Reaction of lead(II) sulfide with ozone

The formula for lead(II) sulfide is \(\mathrm{PbS}\) and reacts with ozone (\(\mathrm{O_3}\)) to form lead(II) sulfate (\(\mathrm{PbSO_4}\)) and oxygen (\(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{PbS} + \mathrm{O_3} \rightarrow \mathrm{PbSO_4} + \mathrm{O_2}$$ To balance the equation, we need 2 \(\mathrm{O_3}\) molecules to produce 2 \(\mathrm{O_2}\) molecules: $$\mathrm{PbS} + 2\mathrm{O_3} \rightarrow \mathrm{PbSO_4} + 2\mathrm{O_2}$$
04

(d) Conversion of zinc sulfide to zinc oxide in air

The formula for zinc sulfide is \(\mathrm{ZnS}\) and is converted to zinc oxide (\(\mathrm{ZnO}\)) when heated in air (in the presence of oxygen, \(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{ZnS} + \mathrm{O_2} \rightarrow \mathrm{ZnO}$$ To balance the equation, we need 1 \(\mathrm{O_2}\) molecule to react with 1 \(\mathrm{ZnS}\): $$\mathrm{ZnS} + \frac{1}{2}\mathrm{O_2} \rightarrow \mathrm{ZnO}$$
05

(e) Reaction of potassium peroxide with carbon dioxide

The formula for potassium peroxide is \(\mathrm{K_2O_2}\) and reacts with carbon dioxide (\(\mathrm{CO_2}\)) to form potassium carbonate (\(\mathrm{K_2CO_3}\)) and oxygen (\(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{K_2O_2} + \mathrm{CO_2} \rightarrow \mathrm{K_2CO_3} + \mathrm{O_2}$$ Since all elements are already balanced, no additional coefficients are needed.
06

(f) Conversion of oxygen to ozone in the upper atmosphere

The conversion of oxygen (\(\mathrm{O_2}\)) to ozone (\(\mathrm{O_3}\)) in the upper atmosphere can be represented by the equation: $$\mathrm{O_2} \rightarrow \mathrm{O_3}$$ To balance the equation, note that 3 \(\mathrm{O_2}\) molecules are needed to form 2 \(\mathrm{O_3}\) molecules: $$\frac{3}{2}\mathrm{O_2} \rightarrow \mathrm{O_3} $$

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Most popular questions from this chapter

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) \(\mathrm{HNO}_{2},(\mathbf{b}) \mathrm{N}_{3}^{-},(\mathrm{c}) \mathrm{N}_{2} \mathrm{H}_{5}{ }^{+},(\mathrm{d}) \mathrm{NO}_{3}^{-}\).

A friend tells you that the "neon" in neon signs is a compound of neon and aluminum. Can your friend be correct? Explain.

Explain each of the following observations: (a) At room temperature \(\mathrm{I}_{2}\) is a solid, \(\mathrm{Br}_{2}\) is a liquid, and \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) are both gases. (b) \(\mathrm{F}_{2}\) cannot be prepared by electrolytic oxidation of aqueous \(\mathrm{F}^{-}\) solutions. (c) The boiling point of \(\mathrm{HF}\) is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\)

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the hightemperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\)

The solubility of \(\mathrm{Cl}_{2}\) in \(100 \mathrm{~g}\) of water at \(\mathrm{STP}\) is \(310 \mathrm{~cm}^{3}\). Assume that this quantity of \(\mathrm{Cl}_{2}\) is dissolved and equilibrated as follows: $$ \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q) $$ (a) If the equilibrium constant for this reaction is \(4.7 \times 10^{-4}\), calculate the equilibrium concentration of \(\mathrm{HClO}\) formed. \((\mathbf{b})\) What is the \(\mathrm{pH}\) of the final solution?
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