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Worked Example 5.1 Balancing Chemical Reactions

Worked Example 5.1 Balancing Chemical Reactions. Use words to explain the following equation for the reaction used in extracting lead metal from its ores. Show that the equation is balanced. Solution.

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Worked Example 5.1 Balancing Chemical Reactions

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  1. Worked Example 5.1 Balancing Chemical Reactions Use words to explain the following equation for the reaction used in extracting lead metal from its ores. Show that the equation is balanced. Solution The equation can be read as, “Solid lead (II) sulfide plus gaseous oxygen yields solid lead (II) oxide plus gaseous sulfur dioxide.” To show that the equation is balanced, count the atoms of each element on each side of the arrow: On the left: 2 Pb 2 S On the right: 2 Pb 2 S The numbers of atoms of each element are the same in the reactants and products, so the equation is balanced.

  2. Worked Example 5.2 Balancing Chemical Equations Write a balanced chemical equation for the Haber process, an important industrial reaction in which elemental nitrogen and hydrogen combine to form ammonia. Solution STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. By examination, we see that only two elements, N and H, need to be balanced. Both these elements exist in nature as diatomic gases, as indicated on the reactant side of the unbalanced equation. STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Remember that the subscript 2 in and indicates that these are diatomic molecules (that is, 2 N atoms or 2 H atoms per molecule). Since there are 2 nitrogen atoms on the left, we must add a coefficient of 2 in front of the on the right side of the equation to balance the equation with respect to N: Now we see that there are 2 H atoms on the left, but 6 H atoms on the right. We can balance the equation with respect to hydrogen by adding a coefficient of 3 in front of the (g) on the left side:

  3. STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. On the left: (1 × 2) N = 2 N (3 × 2) H = 6 H On the right: (2 × 1) N = 2 N (2 × 3) H = 6 H STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, the coefficients already represent the lowest whole-number values.

  4. Worked Example 5.3 Balancing Chemical Equations Natural gas (methane, ) burns in oxygen to yield water and carbon dioxide . Write a balanced equation for the reaction. Solution STEP 1: Write the unbalanced equation, using correct formulas for all substances: (Unbalanced) STEP 2: Since carbon appears in one formula on each side of the arrow, let us begin with that element. In fact, there is only 1 carbon atom in each formula, so the equation is already balanced for that element. Next, note that there are 4 hydrogen atoms on the left (in ) and only 2 on the right (in ). Placing a coefficient of 2 before gives the same number of hydrogen atoms on both sides: (Balanced for C and H) Finally, look at the number of oxygen atoms. There are 2 on the left (in ) but 4 on the right (2 in and 1 in each ). If we place a 2 before the , the number of oxygen atoms will be the same on both sides, but the numbers of other elements will not change: (Balanced for C, H, and O)

  5. STEP 3: Check to be sure the numbers of atoms on both sides are the same. On the left:1 C 4 H On the right:1 C STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, the answer is already correct.

  6. Worked Example 5.4 Balancing Chemical Equations Sodium chlorate decomposes when heated to yield sodium chloride and oxygen, a reaction used to provide oxygen for the emergency breathing masks in airliners. Write a balanced equation for this reaction. Solution STEP 1: The unbalanced equation is: STEP 2: Both the Na and the Cl are already balanced, with only one atom of each on the left and right sides of the equation. There are 3 O atoms on the left, but only 2 on the right. The O atoms can be balanced by placing a coefficient of 1½ in front of on the right side of the equation: STEP 3:Checking to make sure the same number of atoms of each type occurs on both sides of the equation, we see 1 atom of Na and Cl on both sides, and 3 O atoms on both sides. STEP 4: In this case, obtaining all coefficients in their smallest whole-number values requires that we multiply all coefficients by 2 to obtain: Checking gives On the left: On the right: The oxygen in emergency breathing masks comes from heating sodium chlorate.

  7. Worked Example 5.5 Classifying Chemical Reactions Classify the following as a precipitation, an acid–base neutralization, or a redox reaction. (a) (b) (c) Analysis One way to identify the class of reaction is to examine the products that form and match them with the descriptions for the types of reactions provided in this section. By a process of elimination, we can readily identify the appropriate reaction classification. Solution (a) The products of this reaction are water and an ionic compound, or salt . This is consistent with the description of an acid–base neutralization reaction. (b) This reaction involves two aqueous reactants, and NaCl, which combine to form a solid product, . This is consistent with a precipitation reaction. (c) The products of this reaction are a solid, Ag (s), and an aqueous ionic compound, . This does not match the description of a neutralization reaction, which would form water and an ionic compound. One of the products is a solid, but the reactants are not both aqueous compound; one of the reactants is also a solid (Cu). Therefore, this reaction would not be classified as a precipitation reaction. By the process of elimination, then, it must be a redox reaction.

  8. Worked Example 5.6 Chemical Reactions: Solubility Rules Will a precipitation reaction occur when aqueous solutions of and are mixed? Solution Identify the two potential products, and predict the solubility of each using the guidelines in the text. In this instance, and might give CdS and . Since the guidelines predict that CdS is insoluble, a precipitation reaction will occur:

  9. Worked Example 5.7 Chemical Reactions: Acid–Base Neutralization Write an equation for the neutralization reaction of aqueous and aqueous . Solution The reaction of with involves the combination of a proton from the acid with from the base to yield water and a salt .

  10. Worked Example 5.8 Chemical Reactions: Redox Reactions For the following reactions, indicate which atom is oxidized and which is reduced, based on the definitions provided in this section. Identify the oxidizing and reducing agents. (a) (b) Analysis The definitions for oxidation include a loss of electrons, an increase in charge, and a gain of oxygen atoms; reduction is defined as a gain of electrons, a decrease in charge, and a loss of oxygen atoms. Solution (a) In this reaction, the charge on the Cu atom increases from 0 to 2+. This corresponds to a loss of 2 electrons. The Cu is therefore oxidized and acts as the reducing agent. Conversely, the ion undergoes a decrease in charge from 2+ to 0, corresponding to a gain of 2 electrons for the ion. The is reduced, and acts as the oxidizing agent. (b) In this case, the gain or loss of oxygen atoms is the easiest way to identify which atoms are oxidized and reduced. The Mg atom is gaining oxygen to form MgO; therefore, the Mg is being oxidized and acts as the reducing agent. The C atom in is losing oxygen. Therefore, the C atom in is being reduced, and so acts as the oxidizing agent.

  11. Worked Example 5.9 Chemical Reactions: Identifying Oxidizing/Reducing Agents For the respiration and metallurgy examples discussed previously, identify the atoms being oxidized and reduced, and label the oxidizing and reducing agents. Analysis Again, using the definitions of oxidation and reduction provided in this section, we can determine which atom (s) are gaining/losing electrons or gaining/losing oxygen atoms. Solution Respiration: Because the charge associated with the individual atoms is not evident, we will use the definition of oxidation/reduction as the gaining/losing of oxygen atoms. In this reaction, there is only one reactant besides oxygen , so we must determine which atom in the compound is changing. The ratio of carbon to oxygen in is 1:1, while the ratio in is 1:2. Therefore, the C atoms are gaining oxygen and are oxidized; the is the reducing agent and is the oxidizing agent. Note that the ratio of hydrogen to oxygen in and in is 2:1. The H atoms are neither oxidized nor reduced. Metallurgy: The is losing oxygen to form Fe (s); it is being reduced and acts as the oxidizing agent. In contrast, the CO is gaining oxygen to form ; it is being oxidized and acts as the reducing agent.

  12. Worked Example 5.10 Chemical Reactions: Identifying Redox Reactions For the following reactions, identify the atom (s) being oxidized and reduced: (a) (b) Analysis Again, there is no obvious increase or decrease in charge to indicate a gain or loss of electrons. Also, the reactions do not involve a gain or loss of oxygen. We can, however, evaluate the reactions in terms of the typical behavior of metals and nonmetals in reactions. Solution (a) In this case, we have the reaction of a metal (Al) with a nonmetal ( ). Because metals tend to lose electrons and nonmetals tend to gain electrons, we can assume that the Al atom is oxidized (loses electrons) and the is reduced (gains electrons). (b) The carbon atom is the less electronegative element (farther to the left) and is less likely to gain an electron. The more electronegative element (Cl) will tend to gain electrons (be reduced).

  13. Worked Example 5.11 Redox Reactions: Oxidation Numbers What is the oxidation number of the titanium atom in ? Name the compound using a Roman numeral (Section 3.10). Solution Chlorine, a reactive nonmetal, is more electronegative than titanium and has an oxidation number of –1. Because there are 4 chlorine atoms in , the oxidation number of titanium must be +4. The compound is named titanium (IV) chloride. Note that the Roman numeral IV in the name of this molecular compound refers to the oxidation number +4 rather than to a true ionic charge.

  14. Worked Example 5.12 Redox Reactions: Identifying Redox Reactions Use oxidation numbers to show that the production of iron metal from its ore ( ) by reaction with charcoal (C) is a redox reaction. Which reactant has been oxidized, and which has been reduced? Which reactant is the oxidizing agent, and which is the reducing agent? Solution The idea is to assign oxidation numbers to both reactants and products and see if there has been a change. In the production of iron from , the oxidation number of Fe changes from +3 to 0, and the oxidation number of C changes from 0 to +4. Iron has thus been reduced (decrease in oxidation number), and carbon has been oxidized (increase in oxidation number). Oxygen is neither oxidized nor reduced because its oxidation number does not change. Carbon is the reducing agent, and is the oxidizing agent.

  15. Worked Example 5.13 Chemical Reactions: Net Ionic Reactions Write balanced net ionic equations for the following reactions: (a) (b) (c) Solution (a) The solubility guidelines discussed in Section 5.4 predict that a precipitate of insoluble AgCl forms when aqueous solutions of and are mixed. Writing all the ions separately gives an ionic equation, and eliminating spectator ions and gives the net ionic equation. Ionic equation: Net ionic equation: The coefficients can all be divided by 2 to give: Net ionic equation: A check shows that the equation is balanced for atoms and charge (zero on each side).

  16. (b) Allowing the acid HCl to react with the base leads to a neutralization reaction. Writing the ions separately, and remembering to write a complete formula for water, gives an ionic equation. Then eliminating the spectator ions and dividing the coefficients by 2 gives the net ionic equation. Ionic equation: Net ionic equation: A check shows that atoms and charges are the same on both sides of the equation.

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