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Chapter 22: Problem 80

Write a balanced equation for each of the following reactions: (a) Diborane reacts with water to form boric acid and molecular hydrogen. (b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid. (c) Boron oxide dissolves in water to give a solution of boric acid.

Short Answer

Expert verified
a) Balancing the equation for the reaction of diborane with water: \(B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 3H_2\) b) Balancing the equation for the condensation reaction of boric acid: \(4H_3BO_3 \rightarrow H_2B_4O_7 + 5H_2O\) c) Balancing the equation for the reaction of boron oxide with water: \(B_2O_3 + 3H_2O \rightarrow 2H_3BO_3\)

Step by step solution

01

Identify the compounds

Diborane: B2H6, Water: H2O, Boric acid: H3BO3, Molecular hydrogen: H2
02

Write the unbalanced equation

B2H6 + H2O -> H3BO3 + H2
03

Balance the equation

To balance the equation, we adjust the stoichiometric coefficients: B2H6 + 6H2O -> 2H3BO3 + 3H2 b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid.
04

Identify the compounds

Boric acid: H3BO3, Tetraboric acid: H2B4O7
05

Write the unbalanced equation

H3BO3 -> H2B4O7
06

Balance the equation

To balance the equation, we adjust the stoichiometric coefficients: 4H3BO3 -> H2B4O7 + 5H2O c) Boron oxide dissolves in water to give a solution of boric acid.
07

Identify the compounds

Boron oxide: B2O3, Water: H2O, Boric acid: H3BO3
08

Write the unbalanced equation

B2O3 + H2O -> H3BO3
09

Balance the equation

To balance the equation, we adjust the stoichiometric coefficients: B2O3 + 3H2O -> 2H3BO3

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Most popular questions from this chapter

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 \(\mathrm{GeO}_{2}\).

Complete and balance the following equations: (a) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow\) (c) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{3}(a q)+\mathrm{H}^{+}(a q) \longrightarrow\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow\) Which ones of these are redox reactions?

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group \(6 \mathrm{~A}\) element in each: (a) sulfur tetrachloride, (b) selenium trioxide, (c) sodium thiosulfate, (d) hydrogen sulfide, (e) sulfuric acid, (f) sulfur dioxide, (g) mercury telluride.

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

Complete and balance the following equations: (a) \(\mathrm{ZnCO}_{3}(s) \stackrel{\Delta}{\longrightarrow}\) (b) \(\mathrm{BaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) (d) \(\mathrm{CS}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) (e) \(\mathrm{Ca}(\mathrm{CN})_{2}(s)+\operatorname{HBr}(a q)\)
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