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Chapter 22: Problem 70

Write a balanced equation for each of the following reactions: (a) Burning magnesium metal in a carbon dioxide atmosphere reduces the \(\mathrm{CO}_{2}\) to carbon. (b) In photosynthesis, solar energy is used to produce glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and \(\mathrm{O}_{2}\) from carbon dioxide and water. (c) When carbonate salts dissolve in water, they produce basic solutions.

Short Answer

Expert verified
(a) Mg + CO鈧� 鈫� 2MgO + C (b) 6CO鈧� + 12H鈧侽 鈫� C鈧咹鈧佲倐O鈧� + 6O鈧� (c) Na鈧侰O鈧� + H鈧侽 鈫� 2NaOH + H鈧侰O鈧�

Step by step solution

01

Write the unbalanced equation

To write an unbalanced equation, we list the reactants and products involved in the reaction. For this reaction, we have magnesium metal (Mg) reacting with carbon dioxide (CO鈧�) to form magnesium oxide (MgO) and carbon (C). Unbalanced equation: Mg + CO鈧� 鈫� MgO + C
02

Balance the equation

To balance the equation, we compare the number of atoms of each element on both sides and adjust the coefficients accordingly. Here, we have one oxygen on the product side and two oxygens on the reactant side, so we need to add a coefficient of 2 in front of MgO. Balanced equation: Mg + CO鈧� 鈫� 2MgO + C #b) Photosynthesis reaction#
03

Write the unbalanced equation

In the photosynthesis reaction, carbon dioxide (CO鈧�) reacts with water (H鈧侽) to form glucose (C鈧咹鈧佲倐O鈧�) and oxygen gas (O鈧�). Unbalanced equation: CO鈧� + H鈧侽 鈫� C鈧咹鈧佲倐O鈧� + O鈧�
04

Balance the equation

To balance the equation, we adjust the coefficients of each part accordingly to have the same number of atoms of each element on both sides. Balanced equation: 6CO鈧� + 12H鈧侽 鈫� C鈧咹鈧佲倐O鈧� + 6O鈧� #c) Carbonate salts dissolving in water#
05

Write the unbalanced equation

When carbonate salts (such as sodium carbonate, Na鈧侰O鈧�) dissolve in water (H鈧侽), they produce a basic solution (such as sodium hydroxide, NaOH) and carbonic acid (H鈧侰O鈧�). Unbalanced equation: Na鈧侰O鈧� + H鈧侽 鈫� NaOH + H鈧侰O鈧�
06

Balance the equation

To balance the equation, we adjust the coefficients of each part accordingly. Balanced equation: Na鈧侰O鈧� + H鈧侽 鈫� 2NaOH + H鈧侰O鈧�

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Most popular questions from this chapter

Although the \(\mathrm{ClO}_{4}^{-}\)and \(\mathrm{IO}_{4}^{-}\)ions have been known for a long time, \(\mathrm{BrO}_{4}^{-}\)was not synthesized until 1965 . The ion was synthesized by oxidizing the bromate ion with xenon difluoride, producing xenon, hydrofluoric acid, and the perbromate ion. (a) Write the balanced equation for this reaction. (b) What are the oxidation states of Br in the Br-containing species in this reaction?

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the high-temperature 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 \(\mathrm{GeO}_{2}\).

Name the following compounds and assign oxidation states to the halogens in them: (a) \(\mathrm{Fe}\left(\mathrm{ClO}_{3}\right)_{3}\), (b) \(\mathrm{HClO}_{2}\), (c) \(\mathrm{XeF}_{60}\) (d) \(\mathrm{BrF}_{5}\), (e) \(\mathrm{XeOF}_{4}\), (f) \(\mathrm{HIO}_{3}\),

Write a balanced equation for each of the following reactions: (a) hydrolysis of \(\mathrm{PCl}_{5}\) (b) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, (c) reaction of \(\mathrm{P}_{4} \mathrm{O}_{10}\) with water. Carbon, the Other Group 4A Elements, and Boron (Sections 22.9, 22.10, and 22.11)

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 $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{-1}^{0} \mathrm{e} \text { or } \quad{ }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+\beta^{-} $$
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