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Chapter 18: Problem 33

Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is \(\begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow & \mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned}\) Balance this equation.

Short Answer

Expert verified
The balanced chemical equation is: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+4\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\)

Step by step solution

01

Identify the species in the given equation

: Reactants: NaCl (aq), H2SO4 (aq), and MnO2 (s) Products: MnCl2 (aq), H2O (l), and Cl2 (g)
02

Balance the chlorine atoms

: The chlorine atom occurs as follows in the equation: Reactant side: NaCl Product side: MnCl2, Cl2 There is only 1 chlorine atom on the left side, while there are 3 chlorine atoms on the right side. To balance the chlorines, we add a coefficient of 2 before NaCl: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\) Now, we have 2 chlorine atoms on each side of the equation.
03

Balance the hydrogen atoms

: The hydrogen atom occurs as follows in the equation: Reactant side: H2SO4 Product side: H2O There are 2 hydrogen atoms on the left side and 2 hydrogen atoms on the right side. So the hydrogen atoms balanced.
04

Balance the sodium atoms

: The sodium atom occurs as follows in the equation: Reactant side: 2NaCl (due to the coefficient of 2) Product side: none To balance the sodium atoms, there's no need to make any changes as there are 2 sodium atoms on both sides of the equation.
05

Balance the manganese atoms

: The manganese atom occurs as follows in the equation: Reactant side: MnO2 Product side: MnCl2 There is 1 manganese atom on both sides of the equation. The manganese atoms are already balanced.
06

Balance the oxygen atoms

: The oxygen atom occurs as follows in the equation: Reactant side: H2SO4, MnO2 Product side: H2O There are 4 + 2 = 6 oxygen atoms on the left side (4 in H2SO4 and 2 in MnO2) and 1 oxygen atom on the right side (H2O). To balance the oxygen atoms, add a coefficient of 4 before H2O on the product side: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+4\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\) Now, we have 6 oxygen atoms on both sides of the equation.
07

Check the balanced equation

: Reactants: 2 Na, 6 Cl, 2 H, 1 Mn, 4 O Products: 2 Na, 6 Cl, 2 H, 1 Mn, 4 O The balanced chemical equation is: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+4\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\)

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Key Concepts

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

Chemical Reaction
A chemical reaction is a process where reactants are transformed into new substances, known as products. During this transformation, chemical bonds are broken and new bonds are formed. The chlorine gas preparation you've encountered in your exercise is a classic example of a chemical reaction, where chlorine gas is produced through the interaction of sodium chloride, sulfuric acid, and manganese(IV) oxide.

To understand a chemical reaction, it's crucial to look at the substances involved before and after the reaction. For our exercise, the reactants are NaCl (aqueous), H2SO4 (aqueous), and MnO2 (solid), while the products are MnCl2 (aqueous), H2O (liquid), and Cl2 (gas). The balancing of this equation requires careful accounting of each atom type on both sides of the equation to ensure they are equal, reflecting the conservation of matter.
Stoichiometry
Stoichiometry is the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's based on the principle that matter cannot be created or destroyed in chemical reactions (the law of conservation of mass). When resolving stoichiometric problems, such as balancing the provided chemical equation, one must ensure that the number of atoms for each element involved in the reaction is the same on both sides of the equation.

In our exercise, stoichiometry was used to determine the correct coefficients for reactants and products. This ensures that the amounts of sodium, chlorine, hydrogen, manganese, and oxygen atoms are conserved from reactants to products. The balancing steps you went through systematically adjusted coefficients to achieve this conservation. The step-by-step process exemplifies stoichiometrical calculations that are essential in predicting yields of reactions and understanding the proportions of substances required.
Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are a subset of chemical reactions involving the transfer of electrons between two species. Oxidation is the loss of electrons, while reduction is the gain of electrons. These reactions play a significant role in chemistry, including biological processes, energy storage, and corrosion.

In the context of our exercise, the preparation of chlorine gas can be viewed as a redox reaction. Manganese(IV) oxide (MnO2) acts as an oxidizing agent and loses electrons (is reduced) while sodium chloride (NaCl) is the source of chlorine which is oxidized to form chlorine gas (Cl2). The successful balancing of the equation reflects the nuances of redox balancing, where attention must be paid to both mass and charge conservation. Balancing redox reactions involves additional steps compared to other reactions because both the number of atoms and the charges must balance, which adds another layer of complexity to the stoichiometric calculations.

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Most popular questions from this chapter

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M\). Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{~V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

It takes \(15 \mathrm{kWh}\) (kilowatt-hours) of electrical energy to produce \(1.0 \mathrm{~kg}\) aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt \(1.0 \mathrm{~kg}\) aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is \(10.7 \mathrm{~kJ} / \mathrm{mol}(1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .]\)

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.957 \mathrm{~V}\) \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.775 \mathrm{~V}\) a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
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