Solution Manual for General Organic and Biological Chemistry 2nd
Edition by Janice Gorzynski Smith ISBN 0073402788
9780073402789
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Chapter 5 Chemical Reactions
Solutions to In-Chapter Problems
5.1 The process is a chemical reaction because the reactants contain two gray spheres joined (indicating H2) and two red spheres joined (indicating O2), while the product (H2O) contains a red sphere joined to two gray spheres (indicating O–H bonds).
5.2 The process is a physical change (freezing) since the particles in the reactants are the same as the particles in the products.
5.3 Chemical equations are written with the reactants on the left and the products on the right separated by a reaction arrow.
5.4 To determine the number of each type of atom when a formula has both a coefficient and a subscript, multiply the coefficient by the subscript
For 3 Al2(SO4)3: Al = 6 (3 2), S = 9 (3 3), O = 36 (3 3 4)
5.5 Write the chemical equation for the statement.
5.6 Balance the equation with coefficients one element at a time to have the same number of atoms on each side of the equation. Follow the steps in Example 5.2.
5.7 Write the balanced chemical equation for carbon monoxide and oxygen reacting to form carbon dioxide. The smallest set of whole numbers must be used.
5.8 Follow the steps in Example 5.2 to write the balanced chemical equation.
5.9 Write a balanced equation for the Haber process.
5.10 Balance the equations as in Example 5.2.
5.11 One mole, abbreviated as mol, always contains an Avogadro’s number of particles (6.02 1023).
a, b, c, d: 6.02 1023
5.12 Multiply the number of moles by Avogadro’s number to determine the number of atoms. Avogadro’s number is the conversion factor that relates moles to molecules, as in Example 5.3.
a.
5.13 Multiply the number of moles by Avogadro’s number to determine the number of molecules, as in Example 5.3.
a. 2.5 mol 6.02 1023 molecules/mol = 1.5 1024 molecules
b. 0.25 mol 6.02 1023 molecules/mol = 1.5 1023 molecules
c. 0.40 mol 6.02 1023 molecules/mol = 2.4 1023 molecules
d. 55.3 mol 6.02 1023 molecules/mol = 3.33 1025 molecules
5.14 Use Avogadro’s number as a conversion factor to relate molecules to moles.
5.15 To calculate the formula weight, multiply the number of atoms of each element by the atomic weight and add the results.
a.
b.
5.16 Calculate the molecular weight in two steps:
[1] Write the correct formula and determine the number of atoms of each element from the subscripts.
[2] Multiply the number of atoms of each element by the atomic weight and add the results.
5.17
1 Cl atom 35.45 amu = 35.45 amu
3 F atoms 19.00 amu = 57.00 amu
Molecular weight of halothane (C2HBrClF3) 197.378 amu rounded to 197.38 amu
C20H24O10 20 C atoms 12.01 amu = 240.2 amu
24 H atoms 1.008 amu = 24.192 amu
10 O atoms 16.00 amu = 160.0 amu
Molecular weight of ginkgolide B: 424.392 amu = 424.4 g/mol
5.18 Convert the moles to grams using the molar mass as a conversion factor.
a. 0.500 mol of NaCl 58.44 g/mol = 29.2 g
b. 2.00 mol of KI 166.0 g/mol = 332 g
c. 3.60 mol of C2H4 28.05 g/mol = 101 g
d. 0.820 mol of CH4O 32.04 g/mol = 26.3 g
5.19 Convert grams to moles using the molar mass as a conversion factor.
5.20 Use conversion factors to determine the number of molecules in 1.00 g; two 500.-mg tablets = 1.00 g.
5.21 Use mole–mole conversion factors as in Example 5.5 and the equation below to solve the problems.
b. 0.50 mol O2 (2 mol NO/1 mol O2) = 1.0 mol NO
c. 1.2 mol N2 (1 mol O2/1 mol N2) = 1.2 mol O2
5.22 Use mole–mole conversion factors as in Example 5.5 and the equation below to solve the problems.
5.23 [1] Convert the number of moles of reactant to the number of moles of product using a mole
mole conversion factor.
[2] Convert the number of moles of product to the number of grams of product using the product’s molar mass.
5.24 Use the steps outlined in Answer 5.23 to answer the questions.
5.25 Use conversion factors to solve the problems. Follow the steps in Sample Problem 5.14.
5.26 Use conversion factors to solve the problems. Follow the steps in Sample Problem 5.14.
c.
5.27 [1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor.
[2] Convert the number of moles of product to the number of grams of product the theoretical yield using the product’s molar mass.
5.28 Use the steps in Answer 5.27 to solve the problem.
5.29 Use the steps in Sample Problem 5.17 to answer the questions.
5.30 Use conversion factors to solve the problem. Follow the steps in Answer 5.29.
5.31 To determine the overall percent yield in a synthesis that has more than one step, multiply the percent yield for each step.
a. (0.90)10 100% = 35%
b. (0.80)10 100% = 11% c. 0.50 (0.90)9 100% = 19%
5.32 Use the steps in Sample Problem 5.18 to answer the questions.
5.35 Convert the number of grams of each reactant to the number of moles using molar masses. Since the mole ratio of O2 to N2 is 1:1, the limiting reactant has fewer moles.
5.36 Calculate the number of moles of product formed based on the limiting reactant. Then convert moles to grams using molar mass.
5.37
5.38 A compound that gains electrons is reduced. A compound that loses electrons is oxidized.
5.39 A compound that gains electrons while causing another compound to be oxidized is called an oxidizing agent. A compound that loses electrons while causing another compound to be reduced is called a reducing agent.
a. Zn reducing agent, H+ oxidizing agent
b. Fe3+ oxidizing agent, Al reducing agent
c. I– reducing agent, Br2 oxidizing agent
d. Br– reducing agent, Ag+ oxidizing agent
5.40
5.41 H2 is oxidized since it gains an O atom and C2H4O2 is reduced since it gains hydrogen.
5.42 Zn is the reducing agent and Hg2+ is the oxidizing agent.
Solutions to End-of-Chapter Problems
5.43 The process is a chemical reaction because the spheres in the reactants are joined differently than the spheres in the products.
5.44 a. The transformation of [1] to [2] is a chemical reaction because the spheres in the reactants (AB) are joined differently than the spheres in the products (A2 and B2).
b. The transformation of [1] to [3] is a physical change because the spheres are joined the same (AB) but they are now closer together indicating a physical state change from gas to liquid.
5.45 The difference between a coefficient and a subscript is that the coefficient indicates the number of molecules or moles undergoing reaction, whereas the subscript indicates the number of atoms of each element in a chemical formula.
5.46 It is not possible to change the subscripts of a chemical formula to balance an equation because changing the subscripts changes the identity of the compound.
5.47 Add up the number of atoms on each side of the equation and then label the equations as balanced or not balanced.
5.48 Add up the number of atoms on each side of the equation and then label the equations as balanced or not balanced.
5.49 Write the balanced equation using the colors of the spheres to identify the atoms (gray = hydrogen and green = chlorine).
5.50 Write the balanced equation using the colors of the spheres to identify the atoms (red = oxygen and blue = nitrogen).
5.51 Balance the equation with coefficients one element at a time to have the same number of atoms on each side of the equation. Follow the steps in Example 5.2.
5.52 Balance the equation with coefficients one element at a time so that there are the same numbers of atoms on each side of the equation. Follow the steps in Example 5.2.
5.53 Follow the steps in Example 5.2 and balance the equations.
5.54 Follow the steps in Example 5.2 and balance the equation.
5.55 Follow the steps in Example 5.2 and balance the equation.
5.56 Follow the steps in Example 5.2 and balance the equation.
5.57 Fill in the molecules of the products using the balanced equation and following the law of conservation of mass. Each side must have the same number of O and C atoms.
5.58 Fill in the molecules of the products using the balanced equation and following the law of conservation of mass. Each side must have the same number of C, O, and N atoms.
5.59 To calculate the formula weight, multiply the number of atoms of each element by the atomic weight and add the results. The formula weight in amu is equal to the molar mass in g/mol.
5.60 To calculate the formula weight, multiply the number of atoms of each element by the atomic weight and add the results. The formula weight in amu is equal to the molar mass in g/mol.
a. 1 Mg atom 24.30 amu = 24.30 amu
1 S atom 32.07 amu = 32.07 amu
4 O atoms 16.00 amu = 64.00 amu
Formula weight of MgSO4 120.37 amu = 120.37 g/mol
b. 3 Ca atoms 40.08 amu = 120.24 amu
2 P atoms 30.97 amu = 61.94 amu
8 O atoms 16.00 amu = 128.00 amu
Formula weight of Ca3(PO4)2 310.18 amu = 310.18 g/mol
c. 16 C atoms 12.01 amu = 192.16 amu
16 H atoms 1.01 amu = 16.16 amu
1 Cl atom 35.45 amu = 35.45 amu
1 N atom 14.01 amu = 14.01 amu
2 O atoms 16.00 amu = 32.00 amu
2 S atoms 32.07 amu = 64.14 amu
Formula weight of C16H16ClNO2S 353.92 amu = 353.92 g/mol
5.61 Determine the molecular formula of L-dopa. Then calculate the formula weight and molar mass as in Answer 5.59.
5.62 Determine the molecular formula of niacin. Then calculate the formula weight and molar mass as in Answer 5.59
5.63 Convert all of the units to moles, and then compare the atomic mass or formula weight to determine the quantity with the larger mass.
a. 1 mol of Fe atoms (55.85 g/mol) < 1 mol of Sn atoms (118.7 g/mol)
b. 1 mol of C atoms (12.01 g/mol) < 6.02 1023 N atoms = 1 mol N atoms (14.01 g/mol)
c. 1 mol of N atoms (14.01 g/mol) < 1 mol of N2 molecules = 2 mol N atoms (28.02 g/mol N2)
d. 1 mol of CO2 molecules (44.01 g/mol) > 3.01 1023 N2O molecules = 0.500 mol N2O (44.02 g/mol N2O) = 22.01 g N2O
5.64 Convert all of the units to moles, and then compare the atomic mass or formula weight to determine the quantity with the larger mass.
a. 1 mol of Si atoms (28.08 g/mol) < 1 mol of Ar atoms (39.95 g/mol)
b. 1 mol of He atoms (4.00 g/mol) > 6.02 1023 H atoms = 1 mol H atoms (1.01 g/mol)
c. 1 mol of Cl atoms (35.45 g/mol) < 1 mol of Cl2 molecules = 2 mol Cl atoms (70.90 g/mol Cl2)
d. 1 mol of C2H4 molecules (28.06 g/mol) > 3.01 1023 C2H4 molecules = 0.500 mol C2H4 (28.06 g/mol C2H4) = 14.03 g C2H4
5.65 Calculate the molar mass of each compound as in Answer 5.59, and then multiply by 5.00 mol.
a. HCl = 182 g c. C2H2 = 130. g
b. Na2SO4 = 710. g d. Al(OH)3 = 390. g
5.66 Calculate the molar mass of each compound as in Answer 5.59, and then multiply by 0.50 mol.
a. NaOH = 20. g
b. CaSO4 = 68 g
c. C3H6 = 21 g
d. Mg(OH)2 = 29 g
5.67 Convert the grams to moles using the molar mass as a conversion factor.
5.68 Convert the grams to moles using the molar mass as a conversion factor.
5.69 Multiply the number of moles by Avogadro’s number to determine the number of molecules, as in Example 5.3. a.
d.
5.70 Use Avogadro’s number to convert the number of molecules to moles.
5.71 Use the molar mass as a conversion factor to convert the moles to grams. Use Avogadro’s number to convert the number of molecules to moles.
5.72 Use the molar mass as a conversion factor to convert the moles to grams. Use Avogadro’s number to convert the number of molecules to moles.
5.73
a. 12.5 moles of O2 are needed to react completely with 5.00 mol of C2H2
5.00 mol C2H2 (5 mol O2/2 mol C2H2) = 12.5 mol O2
b. 12 moles of CO2 are formed from 6.0 mol of C2H2
6.0 mol C2H2 (4 mol CO2/2 mol C2H2) = 12 mol CO2
c. 0.50 moles of H2O are formed from 0.50 mol of C2H2
0.50 mol C2H2 (2 mol H2O/2 mol C2H2) = 0.50 mol H2O
d. 0.40 moles of C2H2 are needed to form 0.80 mol of CO2
0.80 mol CO2 (2 mol C2H2/4 mol CO2) = 0.40 mol C2H2
5.74
a. 3.0 moles of H2O are needed to react completely with 3.0 mol of Na.
3.0 mol Na (2 mol H2O/2 mol Na) = 3.0 mol H2O
b. 0.19 moles of H2 are formed from 0.38 mol of Na.
0.38 mol Na (1 mol H2/2 mol Na) = 0.19 mol H2
c. 1.82 moles of H2 are formed from 3.64 mol of H2O.
3.64 mol H2O (1 mol H2/2 mol H2O) = 1.82 mol H2
5.75 Use conversion factors as in Example 5.6 to solve the problems.
a. 220 g of CO2 are formed from 2.5 mol of C2H2
b. 44 g of CO2 are formed from 0.50 mol of C2H2
c. 4.5 g of H2O are formed from 0.25 mol of C2H2
d. 240 g of O2 are needed to react with 3.0 mol of C2H2
5.76 Use conversion factors as in Example 5.6 to solve the problems.
a. 120 g of NaOH are formed from 3.0 mol of Na.
b. 0.30 g of H2 are formed from 0.30 mol of Na.
c. 3.6 g of H2O are needed to react with 0.20 mol of Na.
5.77 Use the equation to determine the percent yield.
5.78 Use the equation to determine the percent yield.
5.79 Use the following equations to determine the percent yield.
5.80 Use the following equations to determine the percent yield.
5.81
5.85 Use the limiting reactant from Problem 5.83 to determine the amount of product formed. The conversion of moles of limiting reagent to grams of product is combined in a single step.
5.86 Use the limiting reactant from Problem 5.84 to determine the amount of product formed. The conversion of moles of limiting reagent to grams of product is combined in a single step.
5.87
5.88
5.89 A substance that is oxidized loses electrons, whereas an oxidizing agent gains electrons (it is reduced).
5.90 A substance that is reduced gains electrons, whereas a reducing agent loses electrons (it is oxidized).
5.91 The species that is oxidized loses one or more electrons. The species that is reduced gains one or more electrons.
5.92 The species that is oxidized loses one or more electrons. The species that is reduced gains one or more electrons.
5.93 The oxidizing agent gains electrons (it is reduced). The reducing agent loses electrons (it is oxidized).
5.94 The oxidizing agent gains electrons (it is reduced). The reducing agent loses electrons (it is oxidized).
5.95 Acetylene is reduced because it gains hydrogen atoms.
5.96 Cl2 is reduced because it gains electrons.
5.97 Write the balanced equation and the half reactions.
5.98 Write the balanced equation and the half reactions.
5.99 Refer to prior solutions to answer each part.
a. Calculate the molar mass as in Answer 5.59; the molar mass of sucrose = 342.3 g/mol.
b. Follow the steps in Example 5.2.
c. 8 mol of ethanol are formed from 2 mol of sucrose.
d. 10 mol of water are needed to react with 10 mol of sucrose.
e. 101 g of ethanol are formed from 0.550 mol of sucrose.
f. 18.4 g of ethanol are formed from 34.2 g of sucrose.
g. 9.21 g ethanol
h. 13.6%
5.100Refer to prior solutions to answer each part.
a. Calculate the molar mass as in Answer 5.59; the molar mass of diethyl ether = 74.1 g/mol.
b. Follow the steps in Example 5.2.
c. 1 mol of diethyl ether is formed from 2 mol of ethanol.
d. 5 mol of water are formed from 10 mol of ethanol.
e. 20. g of diethyl ether are formed from 0.55 mol of ethanol.
f. 3.70 g of diethyl ether are formed from 4.60 g of ethanol.
g. 1.85 g diethyl ether
h. 97.3%
5.107
a. Calculate the molar mass as in Answer 5.59; the molar mass of DDT = 354.5 g/mol.
b. 18 g of DDT would be formed from 0.10 mol of chlorobenzene.
c. 17.8 g is the theoretical yield of DDT in grams from 11.3 g of chlorobenzene.
d. 84.3%
5.108 Refer to prior solutions to answer each part.
a. Calculate the molar mass as in Answer 5.59; the molar mass of linolenic acid = 278.5 g/mol.
d. 10.2 grams of C18H36O2 will be formed.
5.109 Use conversion factors to answer the questions about dioxin.
5.110 Pb is the reducing agent. It is oxidized from Pb to Pb2+. PbO2 is the oxidizing agent. Pb is reduced from +4 in PbO2 to +2 in PbSO4.