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Chapter 5–1
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
5.15 To calculate the formula weight, multiply the number of atoms of each element by the atomic weight and add the results.
a. 1 Ca atom × 40.08 amu = 40.08 amu 1 C atom × 12.01 amu = 12.01 amu 3 O atoms × 16.00 amu = 48.00 amu Formula weight of CaCO3 100.09 amu b. 1 K atom × 39.10 amu = 39.10 amu 1 I atom × 126.9 amu = 126.9 amu Formula weight of KI 166.00 amu rounded to 166.0 amu
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.
a. 2 C atoms × 12.01 amu = 24.02 amu 6 H atoms × 1.008 amu = 6.048 amu 1 O atom × 16.00 amu = 16.00 amu Molecular weight of ethanol (C2H6O) 46.068 amu rounded to 46.07 amu
b. 6 H atoms × 1.008 amu = 6.048 amu 6 C atoms × 12.01 amu = 72.06 amu 1 O atom × 16.00 amu = 16.00 amu Molecular weight of phenol (C6H6O) 94.108 amu rounded to 94.11 amu
c. 1 H atom × 1.008 amu = 1.008 amu 2 C atom × 12.01 amu = 24.02 amu 1 Br atom × 79.90 amu = 79.90 amu 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
5.17
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 c. 3.60 mol of C2H4 × 28.05 g/mol = 101 g b. 2.00 mol of KI × 166.0 g/mol = 332 g d. 0.820 mol of CH4O × 32.04 g/mol = 26.3 g
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.
2 O3 2 CO2 + 2 O22 CO + (not balanced) 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.
2 HCl(aq) + Ca(s) CaCl2(aq) H2(g)
TiCl4 2 H2O TiO2 HCl
Al(OH)3 H3PO4 AlPO4 3 H2O
+
+
+
+
+
a.
b.
c.
2 H, 2 Cl, 1 Ca: both sides, therefore balanced
1 Ti, 4 Cl, 4 H, 2 O 1 Ti, 1 Cl, 1 H, 2 O
NOT balanced
1 Al, 1 P, 7 O, 6 H: both sides, therefore 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.
3 NO2 + H2O HNO3 2 NO
2 H2S 3 O2 H2O 2 SO2
Ca(OH)2 2 HNO3 2 H2O Ca(NO3)2
+
+
+
+
+
a.
b.
c.
3 N, 7 O, 2H
4 H, 2 S, 6 O 2 H, 2 S, 5 ONOT balanced
1 Ca 8 O, 4 H, 2 N: both sides, therefore balanced
3 N, 5 O, 1 HNOT balanced
5.49 Write the balanced equation using the colors of the spheres to identify the atoms (gray = hydrogen
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.
reactants products
CO2O2
O3 CO
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.
reactants products
CO2N2CO NO
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.
a. 1 Na atom × 22.99 amu = 22.99 amu 1 N atom × 14.01 amu = 14.01 amu 2 O atoms × 16.00 amu = 32.00 amu Formula weight of NaNO2 69.00 amu = 69.00 g/mol
b. 2 Al atom × 26.98 amu = 53.96 amu 3 S atoms × 32.07 amu = 96.21 amu 12 O atoms × 16.00 amu = 192.0 amu Formula weight of Al2(SO4)3 342.17 amu rounded to 342.2 amu = 342.2 g/mol
c. 6 C atom × 12.01 amu = 72.06 amu 8 H atoms × 1.008 amu = 8.064 amu 6 O atoms × 16.00 amu = 96.00 amu Formula weight of C6H8O6 176.124 amu rounded to 176.12 amu = 176.12 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.
C CH
HCH
NH2
CO
OH
CCC
C C
HHO
HO
H HL-dopa
a. molecular formula = C9H11NO4!!b. formula weight = 197.2 amu!c. molar mass = 197.2 g/mol
5.62 Determine the molecular formula of niacin. Then calculate the formula weight and molar mass as
in Answer 5.59
C CO
OH
CCC
N C
HH
H
Hniacin
a. molecular formula = C6H5NO2!!b. formula weight = 123.1 amu!c. molar mass = 123.1 g/mol
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
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 2 H2O(l) 2 NaOH(aq)+2 Na(s) H2(g)+
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.
actual yield (g)=Percent yieldtheoretical yield (g)
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.
2 mol NO2
2 mol NO2.0 mol NO x = 92 g NO2a.
46.01 g NO2
1 mol NO2x
2 mol NO2
1 mol O22.0 mol O2 x = 180 g NO2b. 46.01 g NO2
1 mol NO2x
2 mol NO2
2 mol NO0.500 mol NO x = 23.0 g NO2c.
46.01 g NO2
1 mol NO2x
2 mol NO2
1 mol O20.125 mol O2 x = 11.5 g NO2d.
46.01 g NO2
1 mol NO2x
5.87
1 mol C2H4
28.05 g C2H4
8.00 g C2H4 x = 0.285 mol C2H4
1 mol HCl36.46 g HCl
12.0 g HCl x = 0.329 mol HCl
a.
Since the mole ratio in the balanced equation is 1:1, the reactant with the smaller number of moles is the limiting reactant: C2H4.
b.
1 mol C2H5Cl1 mol C2H4
0.285 mol C2H4 x = 0.285 mol C2H5Clc.
64.51 g C2H5Cl1 mol C2H5Cl
0.285 mol C2H5Cl x = 18.4 g C2H5Cld.
e. = 57.6 % percent yield10.6 g C2H5Cl18.4 g C2H5Cl
x 100%
5.88
1 mol CH4
16.05 g CH4
5.00 g CH4 x = 0.312 mol CH4
1 mol Cl270.90 g Cl2
15.0 g Cl2 x = 0.212 mol Cl2
a.
Since the mole ratio in the balanced equation is 1:2, (2)(0.312 mol) = 0.624 mol Cl2 would be needed to react with all of the CH4. There are only 0.212 mol Cl2, however, so Cl2 is the limiting reactant.
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.
2 Mg + O2 2 MgO
2 Mg2+2 Mg + 4 e– O2 2 O2–+ 4 e–
5.98 Write the balanced equation and the half reactions.
4 Al + 3 O2 2 Al2O3
4 Al3+4 Al + 12 e– 3 O2 6 O2–+ 12 e–
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.
C12H22O11(s) H2O(l) 4 C2H6O(l) 4 CO2(g)+ +
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%
5.100 Refer 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%
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.