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When balancing equations in Chapter 3, we did not mention that reactions must be charge balanced as well as mass

Short Answer

Expert verified
The given equation is \(Na + Cl_2 \rightarrow NaCl\). To balance the charges, first identify the charges of the species involved: \(Na^+\), \(Cl^-\), and the neutral \(Cl_2\). The net charge on both sides of the arrow is 0, which means the charges are already balanced, and there is no need to adjust the coefficients. The final balanced equation is \(Na_{(s)} + Cl_{2_{(s)}} \rightarrow NaCl_{(s)}\).

Step by step solution

01

Identify the charges in the equation

All species involved in the equation can be found in the periodic table. Sodium (Na), as part of group 1, will form a \(+1\) charge ion when it reacts, i.e., \(Na^+\). Chlorine (Cl), as part of group 17, will form a \(-1\) charge ion when it reacts, i.e., \(Cl^-\). The Cl atoms in elemental form are neutral because they are bonded to each other, taking the form of \(Cl_2\), which carry no charge.
02

Balance the charges

Based on the species' charges, the equation is: \(Na + Cl_2 \rightarrow Na^+ + Cl^-\) The net charge on the left side of the arrow is 0 since Na and Cl_2 are neutral. The net charge on the right side of the arrow is also 0 since \(Na^+\) has a \(+1\) charge and \(Cl^-\) has a \(-1\) charge. The net charges on both sides of the reaction are equal, so there is no need to adjust the coefficients in this case.
03

Rewrite the balanced equation with appropriate phases

Now that we know the charges balance, we can rewrite the balanced equation with appropriate phases. Sodium and Chlorine are solids (s) at room temperature, and NaCl is also a solid. The final balanced equation is: \(Na_{(s)} + Cl_{2_{(s)}} \rightarrow NaCl_{(s)}\) By carrying out these steps, we have confirmed that the given equation is not only balanced for mass but also balanced for the net charges on both sides of the reaction.

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Key Concepts

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

Chemical Reaction
Chemical reactions are processes where reactants transform into products through the breaking and forming of chemical bonds. The substances you start with, called reactants, undergo a change during the reaction and become a new substance, known as the product. One essential rule must be followed when writing a chemical equation: the Law of Conservation of Mass. This law states that mass is neither created nor destroyed in a chemical reaction. Hence, the mass of the reactants must equal the mass of the products.

During the balancing of chemical equations, like the one in the textbook exercise (sodium reacting with chlorine to form sodium chloride), it's crucial to ensure that the number of atoms for each element is the same on both sides of the equation. For example, if you have two chlorine atoms on the reactant side, you must also have two chlorine atoms on the product side to maintain this balance. Balancing equations is akin to solving a puzzle, where you adjust coefficients (the numbers placed before the chemical formulas) until you have equal numbers of each type of atom on both sides.
Charge Balance
Along with the Law of Conservation of Mass, charge balance is an equally important principle that must be heeded during chemical reactions, especially in reactions involving ions. Charge balance means that the total charge of the reactants must be equal to the total charge of the products. Ions are atoms or molecules that have lost or gained electrons, giving them a net positive or negative charge.

The textbook exercise highlights that for sodium (Na), which forms a +1 charged ion, the counter ion, chlorine (Cl), forms a -1 charged ion upon reaction. When these two react, they form sodium chloride, which is neutral. It's crucial to ensure that the charges balance to maintain electrical neutrality in the equation. If there were an imbalance, you would have to adjust the coefficients to make sure the number of positive charges equals the number of negative charges, thus maintaining the charge neutrality of the equation.
Stoichiometry
Stoichiometry is the section of chemistry that pertains to calculating the quantities of reactants and products involved in a chemical reaction. It's a concept that lets us predict how much product will form during a reaction or how much of a reactant is required to produce a certain amount of product. Understanding stoichiometry is vital because it underpins the ability to create the exact amount of a substance needed in the industry or to predict the outcome of a reaction in a laboratory.

In the context of the sodium and chlorine reaction from the exercise, stoichiometry would involve determining the mole ratio between the reactants and products, based on the balanced chemical equation. Since the equation is balanced as a one-to-one mole ratio, a mole of sodium reacts with a mole of chlorine gas to produce a mole of sodium chloride. Recognizing and applying stoichiometry is fundamental for a chemist because it elucidates how substances interact with one another quantitatively.

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Most popular questions from this chapter

An aqueous solution of \(\mathrm{PdCl}_{2}\) is electrolyzed for \(48.6\) seconds, and during this time \(0.1064 \mathrm{~g}\) of \(\mathrm{Pd}\) is deposited on the cathode. What is the average current used in the electrolysis?

Consider the following galvanic cell:Calculate the \(K_{\text {sp }}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\). Note that to obtain silver ions in the right compartment (the cathode compartment), excess solid \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) was added and some of the salt dissolved.

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{M} \mathrm{Ag}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\).a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?\)

Consider the following half-reactions: $$ \begin{aligned} \mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{aligned} $$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 M\) in chloride ion and \(0.020 M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 M \mathrm{AlF}_{3}\) solution c. \(1.0 M \mathrm{MnI}_{2}\) solution

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