Chemical Quantities and Aqueous Reactionsptfaculty.gordonstate.edu/lgoodroad/SUMMER 2011... · Chemical Quantities and Aqueous Reactions ... An indicator is a dye whose color depends

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Chemical Quantitiesand AqueousReactions

ew QuestionsReaction stoichiometry is the numerical relationships between chemical amounts in a bal-anced chemical equation. The coefficients in a chemical reaction specify the relative amountsin moles of each of the substances involved in the reaction.

The limiting reactant is the reactant that is completely consumed in a chemical reaction and lim-its the amount of product. The theoretical yield is the amount of product that can be made in achemical reaction based on the amount of limiting reactant. The percent yield is calculated

actual yieldas — x 100. The reactant in excess is any reactant that occurs in a quantity greater

theoretical yieldthan that required to completely react with the limiting reactant. Some of this reactant will be leftover when the reaction is complete.

No, the percent yield would not be different if the actual yield and theoretical yield were cal-culated in moles. The relationship between grams and moles is the molar mass. This wouldbe the same value for the actual yield and the theoretical yield.

An aqueous solution is a solution in which water acts as the solvent. The solvent is the major-ity component of the mixture, and the solute is the minority component in the mixture.

Molarity is a concentration term. It is the amount of solute (in moles) divided by the volumeof solution (in liters). The molarity of a solution can be used as a conversion factor betweenmoles of the solute and liters of the solution.

Substances that completely dissociate into ions when they dissolve in water are called strongelectrolytes and conduct electricity easily. Substances that do not completely dissociate inwater are called weak electrolytes and conduct electricity only weakly. Compounds that donot dissociate into ions when dissolved in water are called non-electrolytes and do not con-duct electricity.

Acids are molecular compounds that ionize—form ions—when they dissolve in water. Astrong acid is one that completely ionizes in solution. A weak acid is one that does not com-pletely ionize in water. A solution of a weak acid is composed mostly of the non-ionized acid.

A compound is termed soluble if it dissolves in water. A compound is insoluble if it does notdissolve in water.

The solubility rules are a set of empirical rules that have been inferred from observationson many ionic compounds. The solubility rules allow us to predict if a compound is solu-ble or insoluble.

Cations that usually form soluble compounds are Li"1', Na+, K+, and NH4+. The anions that usu-

ally form soluble compounds are NO3~" and CaHsCV, which have no exceptions; Cl~, Br~, I~except when these ions pair with Ag+, Hg22+, or Pb2+, which result in insoluble compounds;and SO4 ~ except with Sr^+, Ba , Pb , Ag+ or Ca , which form insoluble compounds. The

115

116 Chapter 4 Chemical Quantities and Aqueous Reactions

anions that usually form insoluble compounds are OH" and S2~ except with Li+, Na+, K+, and NH4+, whichform soluble compounds and when S2~ pairs with Ca2+,Sr2+ or Ba2+ the compounds are soluble; CC>32~ andPO43" are insoluble except when paired with Li+, Na+, K+, and

4.11 A precipitation reaction is one in which a solid or precipitate forms upon mixing two solutions. An exam-ple is 2 Kl(fl^) + Pb(NO3)2(fl<7) -» PbI2(s) + 2 KNO3(fl(j).

4.12 The key to predicting precipitation reactions is to understand that only insoluble compounds form precipi-tates. In a precipitation reaction, two solutions containing soluble compounds combine and an insolublecompound precipitates.

4.13 A molecular equation is an equation showing the complete neutral formulas for each compound in the reac-tion as if they existed as molecules. Equations that list individually all of the ions present as either reactantsor products in a chemical reaction are complete ionic equation. Equations that show only the species thatactually change during the reaction are net ionic equations.

An Arrhenius acid is a substance that produces H+ ions in aqueous solutions. An Arrhenius base is a sub-stance that produces OH~ ions in aqueous solutions.

4.15 When an acid and base are mixed, the H+(a</) from the acid combines with the OH~ from the base to formH2O(/). An example is HClfcq) + NaOH(aq) -» H2O(/) + 1

In a titration, a substance in a solution of known concentration is reacted with another substance in a solu-tion of unknown concentration. The acid-base titration is continued until the neutralization is complete. Atthe equivalence point, the point when the number of moles of OH~ equals the number of moles of H+, thetitration is complete. An indicator is a dye whose color depends on the acidity or basicity of the solution.

4.17 Aqueous reactions that form a gas upon mixing two solutions are called gas-evolution reactions. An exam-ple is H2SO4(fl<7) + U2S(aq) -> H2S(g) + Li2SO4(ag).

4.18 The reactant types that give rise to gas-evolution reactions are sulfides, carbonates, bicarbonate, sulfites,bisulfites, and ammonium compounds.

Oxidation-reduction reactions or redox reactions are reactions in which electrons are transferred from onereactant to the other. An example is 4 Fe(s) + 3 O2(g) —> 2 Fe2O3(s).

i4.20 The oxidation state or oxidation number is a number given to each atom based on the electron assignments.

It is the charge an atom would have if all shared electrons were assigned to the atom with a greater attrac-tion for those electrons.

4.21 To identify redox reactions by using oxidation states, begin by assigning oxidation states to each atom in thereaction. A change in oxidation state for the atoms indicates a redox reaction.

4.22 When a substance is oxidized it loses electrons and there is an increase in oxidation state. When a substanceis reduced it gains electrons and there is a reduction in oxidation state.

4.23 A substance that causes the oxidation of another substance is called an oxidizing agent. A substance thatcauses the reduction of another substance is called a reducing agent.

4.24 Combustion reactions are characterized by the reaction of a substance with O2 to form one or more oxygencontaining compounds, often including water. Combustion reactions emit heat. Combustion reactions areimportant because most of our society's energy is derived from them. An example is Crt}(g) + 2 O2(g) —»C02(g) + 2 .

Chapter 4 Chemical Quantities and Aqueous Reactions 117

Problems by Topic

Reaction Stoichiometry

&

Given: 7.2 moles CgH^ Find: balanced reaction, moles O2 requiredConceptual Plan: balance the equation then mol C^H14 —> mol O2

19molO2

19 mol O2Solution: 7.2 mtJKrgH^ x—————- = 68.4 mol O2 = 68molO2

Check: The units of the answer (mol O2) are correct. The magnitude is reasonable because much more O2 isneeded

4.26 Given: 0.461 moles HC2H3O2 Find: balanced reaction, moles Ba(OH)2 requiredConceptual Plan: balance the reaction then mol HC2H3O2 — * mol Ba(OH)2

1 mol Ba(OH)2 HC2H302 (*?) + Ba(OH)2(fl<f) -» 2 H2O(/) + Ba(C2H3O2)2(atl) 2molHC2H3O2

1 mol Ba(OH)2Solution: 0.461 motHejHA, x - - = 0 .2305 mol Ba(OH)2 = 0.231 mol Ba(OH)2

Check: The units of the answer (mol Ba(OH)2) are correct. The magnitude is reasonable because much lessBa(OH)2 is needed than HC2H3O2.

(a) Given: 2.5 mol N2O5 Find: mol NO2

Conceptual Plan: mol N2Os —* mol NO24NO2

2N2OS

4 mol NO?Solution: 2. 5 mDHÔsx^——- = 5.0molNO2

Check: The units of the answer (mol NO2) are correct. The magnitude is reasonable since it is greaterthan mol N2Os.

(b) Given: 6.8 mol N2O5 Find: mol NO2

Conceptual Plan: mol N2Os — » mol NO24NO2

2N205

4 mo!NO2Solution: 6.8 atDtNzQs x — ^ = 13.6 mol NO2 = 14 mol NO2

Check: The units of the answer (mol NO2) are correct. The magnitude is reasonable since it is greaterthan mol N2Os.

(c) Given: 15.2 g N2O5 Find: mol NO2

Conceptual Plan: g N2O5 -» mol N2O5 -» mol NO21 mol N2O5 4 N02

108.02gN2O5 2N2O5 (

1 fflOrfÔs 4 mol NO2Solution: 15.2 t ^ * * - == 0-2814molNO2 == 0.281 molNO2

Check: The units of the answer (mol NO2) are correct. The magnitude is reasonable since 15 g is about0.13 mol N2O5 and the answer is greater than mol N2O5.

(d) Given: 2.87 kg N2O5 Find: mol NO2

Conceptual Plan: kg N2O5 -» g N2O5 -» mol N2O5 -» mol NO21000 g N2O5 1 mol N2O5 4 NO2

kg N205 108.02gN205 2N2O5


4.28 (a)

(b)

(c)

(d)

Solution:4 mol

= 53.1molNO2

Check: The units of the answer (mol NO2) are correct. The magnitude is reasonable since 2.87 kg asabout 27 mol N2Os and the answer is greater than mol N2Os.

•* --Given: 2.6 mol N2H4 Find: mol NH3

Conceptual Plan: mol N2H4 — * mol NH3

Solution: 2

4NH3

3N^44 molNH3 = 3.46molNH3 = 3.5molNH3

Check: The units of the answer (mol NH3) are correct. The magnitude is reasonable since it is greaterthan mol

Given: 3.55 mol N2H4 Find: mol NH3

Conceptual Plan: mol N2H4 — -> mol NH34NH,

Solution: 3. 554 molNH3

= 4.733molNH3 = 4.73molNH3

Check: The units of the answer (mol NH3) are correct. The magnitude is reasonable since it is greaterthan mol N2Hj.

Given: 65.3 g N2H4 Find: mol NH3

Conceptual Plan: g N2H4 -» mol N2H4 — » mol NH3

4NH,1 mol N2H4

32.05gN2H4

Solution: 65.3

3 N2H4

4 molNH3

32.05= 2.716molNH3 = 2.72molNH3

Check: The units of the answer (mol NH3) are correct. The magnitude is reasonable since there isabout 2 mol N2H4 and the answer is greater than mol N2Hj.

Given: 4.88 kg N2H4 Find: mol NH3

Conceptual Plan: kg N2H4 -» g N2H4 mol N2H4 —> mol NH31000 g N2H4 1 mol N2H4

kgN2H| 32.05gN2H^

4 NH

Solution:

4.88fcg-N2H4x

Check: The units of the answer (mol NH3) are correct. The magnitude is reasonable since 4.88 kg isabout 150 mol N2H4 and the answer is greater than mol!

1 motN^ 4 mol NH3*32nCJTj xo^^T J = 203.0molNH3 = 203molNH3

Given: 3 mol SiO2 Find: mol C, mol SiC, mol COConceptual Plan: mol SiO2 -» mol C -> mol SiC -> mol CO

3_C_ SiC 2 COSiC^ SiO2

Solution: 3 iftoi-SiOg x -£ ^7.7 = 9 mol C 3 mrrf-SiO^ x ——. = 3 mol SiC

2 mol CO

SiO2

3 mol C

Given: 6 mol C Find: mol SiO2, mol SiC, mol COConceptual Plan: mol C —> mol SiO2 -» mol SiC -> mol CO

Si°2 SiC 2 CO3C 3C 3C


Check: The units of the answers (g HBr, g H2) are correct. The magnitude of the answers is reasonablebecause molar mass HBr is greater than Fe and molar mass H2 is much less than Fe.

4.32 Given: 15.2 g Al Find: g H2SO4; g H2

Conceptual Plan: g Al -» mol Al -> mol H2SO4 -> g H2SO4

molAl amolH.SO, 98.09gH2SO4

26 .98 g Al 2 mol Al mol H2SO4

g Al — * mol Al — * mol H2 — * g H2

molAl 3m°1H226 .98 g Al 2 mol Al mol H2

IffloKAl Smot-HzSQ* 98.Solatia 1 5 . 2 ^ 1 x x - 5 l x T ^ r = 82.9gH2SO4

2.016gH2

iôf^ : L70gH2

Check: The units of the answers (g H2SO4/ g H2) are correct. The magnitude of the answers is reasonablebecause molar mass H2SO4 is greater than Al and molar mass H2 is much less than Al.

(a) Given: 3.67 g Ba Find: g BaCl2Conceptual Plan: g Ba -» mol Ba -> mol BaCl2 -> g BaCl2

molBa 1 mol BaCl, 208 .23 g BaCl,137.33 g Ba 1 mol Ba 1 mol BaCl2

1 ffioi-BaG^ 208 .23 g BaCl2- 5.5647gBaQ2 = 5.56gBaCl2

Check: The units of the answer (g BaCl2) are correct. The magnitude of the answer is reasonablebecause it is larger than grams Ba.

(b) Given: 3.67 g CaO Find: g CaCO3

Conceptual Plan: g CaO -* mol CaO -» mol CaCO3 -> g CaCO3

molCaO molCaC03 100.09gCaC03

56 .08 g CaO 1 mol CaO mol CaCOs

Solution:

Check: Units of answer (g CaCO3) are correct. The magnitude of the answer is reasonable because itis larger than grams CaO.

(c) Given: 3.67 g Mg Find: g MgOConceptual Plan: g Mg — * mol Mg — > mol MgO — » g MgO

mol Mg mol MgO 40 .30 g MgO24 .30 g Mg molMg mol MgO

imotMgO 40 .30 g MgO= 6.086gMgO = 6.09gMgO

Check: The units of the answer (g MgO) are correct. The magnitude of the answer is reasonablebecause it is larger than grams Mg.

(d) Given: 3.67 g Al Find: g A12O3

Conceptual Plan: g Al — * mol Al — » mol A12O3 — » g A12O3

molAl 2molA12°3 101.96gAl203

26 .98 g Al 4 mol Al mol A12O3

101 .Solution: 3.67 gA1,_— p - , . ^ . 6.934gA,2Oj . .J

Check: The units of the answer (g A12O3) are correct. The magnitude of the answer is reasonablebecause it is larger than grams Al.

.hapter 4 Chemical Quantities and Aqueous Reactions 129

Given: 4.2 mol ZnS, 6.8 mol O2 Find: Mole amount of excess reactant leftConceptual Plan: mol ZnS — > mol ZnO

2mo'lZr£ ~~* smallest mol amount determines limiting reactant

mol O2 —> mol ZnO2 mol ZnO3 mol O2

then: mol limiting reactant — » mol excess reactant required — * mol excess reactant left2 mol ZnS3 mol O2

Solution: 4. 2 moi-ZfiSx m°, = 4. 2 mol ZnO

0

6.8rrTofQx2m0l,Z"0 = 4.5 mol ZnOSTrtDfQ?

ZnS is the limiting reactant, therefore, O2 is the excess reactant.3 mol O2

4 .2 frioi-ZnS x — ; = 6 . 3 mol O2 required2 -

6.8molO2 -- 6.3molO2 = 0.5 mol O2 leftCheck: The units of the answer (mol O2) are correct and the magnitude is reasonable since it is less than theoriginal amount of O2.

4.44 Given: 0.223 mol FeS, 0.652 mol HC1 Find: Mole amount of excess reactant leftConceptual Plan: mol FeS -* mol FeCl2

1 mol FeCl1 mol FeS ~~* smallest mol amount determines limiting reactant

mol HC1 -» mol FeCl21 mol FeCL2 mol HC1

then: mol limiting reactant —» mol excess reactant required —> mol excess reactant left1 mol FeS2 mol HC11 mol FeCl2

Solution: 0.223 ntcd-FeSx- - = 0.223 mol FeCl2IRtoHFeS

0.652ffloHHax-^—~- = 0.326 mol FeCl22 rrTcrr-HClFeS is the limiting reactant, therefore, HC1 is the excess reactant.

0.223 fftoi-ZPeS x 2 mo1 HC1 = 0.446 mol HC1 requiredlirfDi-FeS

0.652 molHC1 -- 0.446 mol HC1 = 0.206 mol HC1 leftCheck: The units of the answer (mol HC1) are correct and the magnitude is reasonable since it is less thanthe original amount of HC1.

(a) Given: 2.0 g Al, 2.0 g C12 Find: Theoretical yield in g A1C13

Conceptual Plan: g Al mol Al —* mol A1C13

26*98 LAI 2 mol Al3 ~* smallest mol amount determines limiting reactant

g C12 ->• mol C12 -» mol A1C13

ImolCL, 2molAlCl3

70.90gCl2 3molCl2

then: mol A1C13 -» g A1C13

133.3gAlCl3

mol A1C13

130 Chapter 4 Chemical Quantities and Aqueous Reactioi.

2 mol A1C13

133.3gAlCl30.0188 mol AlCl3x- = 2.5gAlCl3mol A1C13

Check: The units of the answer (g A1C13) are correct. The answer is reasonable since Cl2 produced thesmallest amount of product and is the limiting reactant.

(b) Given: 7.5 g Al, 24.8 g C12 Find: Theoretical yield in g A1C13

Conceptual Plan: g Al — > mol Al — > mol A1C13

1 mol Al ^ mo' A1C1.,26.98gAl 2 mol Al ' ~" smallest mol amount determines limiting reactant

g C12 -» mol CI2 -> mol A1C131 mol C12 2 mol A1C13

70.90gCl2 3molCl2

then: mol A1C13 — > g A1C13

molAlClj

2 molSolution: 7.5 g ^ x - x = 0 .2780 mol A1C13

mol- = 0.2332 mol A1C13

133.3gAlCl30.2332 mol A1C13 x - ° = 31.1gAlCl3

Check: The units of the answer (g A1C13) are correct. The answer is reasonable since C12 produced thesmallest amount of product and is the limiting reactant.

(c) Given: 0.235 g Al, 1.15 g C12 Find: Theoretical yield in g A1C13

Conceptual Plan: g Al — > mol Al — > mol A1C13

1 mol Al 2 mo1 A1C1326.98gAl 2 mol A1 — * smallest mol amount determines limiting reactant

g C12 -» mol C12 -> mol A1C13

2molAlCl3

70.90gCl2 3molCl2

then: mol A1C13 -» g A1C13

mol Aids

1 nrol-Al 2 mol A1C13Solution: 0. 235 g-A4x- -x- - = 0 .008710 mol A1C1326.98 g-Al 2mDl-Al

2 mol

133.34gAlCl30.008710 mol A1C13 x - = 1.16gAlCl3mol A1C13

Check: The units of the answer (g A1C13) are correct. The answer is reasonable since Al produced thesmallest amount of product and is the limiting reactant.

4.46 (a) Given: 5.0 g Ti, 5.0 g F2 Find: Theoretical yield inConceptual Plan: g Ti ^ mol Ti — » mol TiF4

i mol Ti4 ~* smallest mol amount determines limiting reactant

g F2 — * mol F2 — >• mol1 mol F2 1 mol TiF4

•^S nn g F2 3 mnl

then: mol TiF4 —» g TiF4

123.87gTiF4

mol TiF4


IT)

Solution: 5.0 i x x = 0.104 mol TiF4

1 mol TiF4

T^Sr = °'°658 mo1 TiF4

123.87gTiF40.0658 mol TiF4 x * = S.lgTiF

Check: The units of the answer (g TiF4) are correct. The answer is reasonable since F2 produced thesmallest amount of product and is the limiting reactant.

(b) Given: 2.4 g Ti, 1.6 g p2 Find: Theoretical yield in g TiF4

Conceptual Plan: g Ti -> mol Ti -» mol TiF4

1 mol TiF47.87gTi i moi Ti ~~* smallest mol amount determines limiting reactant

g F2 —* mol F2 —* mol TiF4

1 mol F2 1 mol TiF4

38.00gF2 2molF2

then: mol TiF4 — » g TiF4

123.87gTiF4

mol TiF4

= 0.0501 mol TiF4

1 mol TiF4

123.87gTiF40.0210 mol TiF4 x - = 2.6gTiF4mol TiF4


(c) Given: 0.233 g Ti, 0.288 g F2 Find: Theoretical yield in g TiF4

Conceptual Plan: g Ti -> mol Ti -» mol TiF41 mol TiF4

47.87 g Ti I mol Ti ~~* smallest mol amount determines limiting reactant

g F2 — > mol F2 -» mol TiF4

1 mol F2 1 mol TiF4

38.00gF2 2molF2

then: mol TiF4 -> g TiF4

123.87gTiF4

mol TiF4

Sdudon, O.

mol Ti

123.87gTiF40.003789 mol TiF4 x - = 0.469gTiF4mol TiF4


Given: 22.55 Fe2O3, 14.78 g CO Find: Mole amount of excess reactant leftConceptual Plan: g Fe2O3 — > mol Fe2O3 — > mol Fe

159 7 g Fe2O3 1 mS°FeÔ3 ~* smallest mol amount determines limiting reactant

g CO —> mol CO — * mol Fe1 mol CO 2 mol Fe

28.01 g CO 3 mol CO


then: mol limiting reactant —» mol excess reactant required —» mol excess reactant left —» g excess1 mol Fe O 159.7 g Fe O 28.oig CO

reactant left 3moico imoiFe^ °rTôfco

1 mol Fe2O3 2 mol Fe

Fe2O3 is the limiting reactant, therefore, CO is the excess reactant.3 frfoi-€O 28.01 g CO

= ^ -^ g CO required

14.78 g CO - 11. 87 g CO = 2.91 g CO left

Check: The units of the answer (g CO) is correct and the magnitude is reasonable since it is less than theoriginal amount of CO.

4.48 Given: 45.69g P4/ 131.3 g C12 Find: Mole amount of excess reactant leftConceptual Plan: g P4 -> mol P4 -> mol PC13

1 mol P. 4 mol PCI.123 88 g P4 I mol P4 ~* smallest mol amount determines limiting reactant

g C12 -* mol C12 -> mol PC13

1 mol Clj 1 mol PC13

70.90 g C12 6 mol C12

then: mol limiting reactant — » mol excess reactant required — * mol excess reactant left — » g excess6molCL 123 . 88 g P. 70.90 gCL

reactant left ^^ imolP« or imolCl.

4molPCl3Solution: 45. 69 enToH^x- — x- — = 1.475 mol PC13~ -

4 mol PC13x- —^ = 1.235molPCl370.90

C12 is the limiting reactant, therefore, P4 is the excess reactant.1 rrrot-€l2 1 BTDi 123.88 g P4

= 38. 236 gP4 required70.90 g-eij 6 ftToi^ls 1 frtol-P^

45.69gP4 - 38.236gP4 = 7.45 gP4 left

Check: Units of the answer (g P4) is correct and the magnitude is reasonable since it is less than the originalamount of P4.

4.49 Given: 28.5 g KC1; 25.7 g Pb2+; 29.4 g PbCl2 Find: limiting reactant, theoretical yield PbCl2, % yieldConceptual Plan: g KC1 -» mol KC1 -» mol PbCl2

1 mol KC"} ^ mol PbCL74 55 g KCl 2 mol KCl ~* smallest mol amount determines limiting reactant

g Pb2+ -» mol Pb2+ -* mol PbCl2

lmolPb2+ 1 mol PbCl2

207.2gPb2+ lmolPb2+

then: mol PbCl2 -» g PbCl2 then: determine % yield278 .1 g PbCl2 actual yield g PbCL,

mol PbCl2 theoretical yield g PbClz X

25 .7 g~Pb2± x 7 x 1 mo1 2 = o .1240 mol PbCl2 Pb2+ is the limiting reactant.207.2 ffl£t 1 2


4.52 Given: 155.8 kg SiO2; 78.3 kg C; 66.1 kg Si Find: limiting reactant, theoretical yield Si, % yieldConceptual Plan: write and balance the reaction, then

kg SiO2 -» g SiO2 -> mol SiO2 — mol Si1000 g 1 mol SiO, | j sj

Tkg~ 60.09gSi02 2 mol SiO2 ~~* sma»est amount determines limiting reactant

kg C -> g C -» mol C -» mol Si1000 g lm°lC02 1 mol Si1kg 44.01 gCOz 2molC

then: mol Si —> gSi—> kg CH4N2O then: determine % yield28.09 g Si 1 kg actual yield kg Sij mo] 5; 1000 g theoretical yield kg Si x 10°

Solution: SiO2(/) + 2C(s) -» Si(/) + 2CO(g)

ISS.Skg-SiQjx — — x — r x — ~ = 2592.8 mol Sikg 60.09g3i02 1 ffrorSiQs

1000 g- imri-C 1 mol Si78.3 te-C x - - x - x - - = 3259.8 mol Si. SiO2 is the limiting reactant

kg 12.01 g-C 2rETDt-C

28.09g-Si kg259?-8m^ X T1JJ X = 72.831 kgSi

11 X 10° = 90'8%

Check: The theoretical yield has the correct units (kg Si) and has a reasonable magnitude compared to themass of SiO2, the limiting reactant. The % yield is reasonable, under 100%.

Sohition Concentration and Solution Stoichiometry

[4.53^ (a) Given: 3.25 mol LiCl; 2.78 L solution Find: Molarity LiClConceptual Plan: mol LiCl, L solution —» Molarity

amount of solute (in moles)molarity (M) = volume of solution (in L)

Solution: 3-25molUC1 = !.169M . 1J7M2.78 L solution

Check: The units of the answer (M) are correct. The magnitude of the answer is reasonable.Concentrations are usually between 0 M and 18 M.

(b) Given: 28.33 g C6H12O6; 1.28 L solution Find: Molarity C6H12O6

Conceptual Plan: g C6H12O6 —» mol C6H12O6, L solution —» Molaritymol C6H]2O6 amount of solute (in moles)

180.16gC6H1206 molarity (M) = volume of soiution (in L)

Solution: 28.33 g-ggHrA, x - = 0.15724 mol C6H12O6BU.lbgCgHTTt^

0.15724 mol C6H12O6- = 0.1228 M = 0.123 M

1.28 L solutionCheck: The units of the answer (M) are correct. The magnitude of the answer is reasonable.Concentrations are usually between 0 M and 18 M.

(c) Given: 32.4 mg NaCl; 122.4 mL solution Find: Molarity NaClConceptual Plan: mg NaCl — > g NaCl — » mol NaCl, and mL solution — > L solution then Molarity

gNaCl mol NaCl L solution . _ amount of solute (in moles)1000 mg NaCl 58 .45 g NaCl 1000 mL solution inty (M) ~ volume of solution (in L)

Solution: 32.4 ntgâd x x = 5.543 x lO^mol NaCl


122.4mt-soh±fenx ^L T = 0.1224L

1000 rn-t,

5.543xlO~4molNaCl- = 0. 0045287 MNaCl = 0. 00453 M NaCl

U. I -- — T 1

Check: The units of the answer (M) are correct. The magnitude of the answer is reasonable.Concentrations are usually between 0 M and 18 M.

4.54 (a) Given: 0.38 mol LiNO3; 6.14 L solution Find: Molarity LiNO3

Conceptual Plan: mol LiNO3, L solution — > Molarityamount of solute (in moles)

molanty (M) = volume of soiution (in L)

0.38 mol LiNO3Solution:- - = 0. 06189 M = 0.062M


(b) Given: 72.8 g C2H6O; 2.34 L solution Find: Molarity C6H12O6

Conceptual Plan: g C2H6O — » mol C2H6O, L solution — » Molaritymol C2H6O amount of solute (in moles)

46 .068 g C2H60 nolanty (M) - volume of soiution (in L)

Solution: 72.8 g - e j H B O x - ^ = 1 .580 mol

1.580 mol C2H6O- = 0.6753 M = 0.675 M


(c) Given: 12.87 mg KI; 112.4 mL solution Find: Molarity KIConceptual Plan: mg KI — » g KI — > mol KI, and mL solution — » L solution then Molarity

;K1 mol KI L solution amount of solute (in moles)I m g K I 166.00gKI 1000 mL solution molanty (M) - volume of solution (in L)

Solution: 1 2 * 7 f t x x = 7.7530 x 10^ mol KI

112.4 mtrsônx = 0.1124 L

7.7530 xlO~5 mol KI= 6.8977xlO"4MKI = 6.898 x 10 ~ 4 M

0.1124 LCheck: The units of the answer (M) are correct. The magnitude of the answer is reasonable.Concentrations are usually between 0 M and 18 M.

(a) Given: 0.556 L; 2.3 M KCI Find: mol KCIConceptual Plan: volume solution x M = mol

volume solution (L) x M = mol

Solution: 0. 556 rSTjhrtion x — — - = 1.3 mol KCI

Check: The units of the answer (mol KCI) are correct. The magnitude is reasonable since it is less than1 L solution.

(b) Given: 1.8 L; 0.85 M KCI Find: mol KCIConceptual Plan: volume solution x M = mol

volume solution (L) x M = mol

Solution: 1 .8 LTsortrtien x - - = 1 .5 mol KCI


Check: The units of the answer (mol KC1) are correct. The magnitude is reasonable since it is less than2 L solution.

(c) Given: 114 mL; 1.85 M KC1 Find: mol KC1Conceptual Plan: mL solution — » L solution, then volume solution x M = mol

1000 mL volume solution (L) x M = mol

Solution: 114 mtrsokrtion x — — rx~iz ' - = 0.211 mol KC11000 mi trsohrtien

Check: The units of the answer (mol KC1) are correct. The magnitude is reasonable since it is less than1 L solution.

4.56 (a) Given: 0.45 mol C2H5OH, 0.200 M C2H5OH Find: volume solutionConceptual Plan: mol C2H5OH — » volume solution

mol C2H5OH

MC2H5OHSolution:

0.200L solution

Check: The units of the answer (L C2H5OH) are correct. The magnitude is reasonable for the amountand volume of solution.

(b) Given: 1.22 mol C2H5OH, 0.200 M C2H5OH Find: volume solutionConceptual Plan: mol C2H5OH — > volume solution

molC2H5OHM C2HsOH

Solution:

0.200L solution

Check: The units of the answer (L C2H5OH) are correct. The magnitude is reasonable for the amountand volume of solution.

(c) Given: 1.2 x 10 ~2 mol C2H5OH/ 0.200 M C2H5OH Find: volume solutionConceptual Plan: mol C2H5OH —> volume solution

molC2H5OHM C2H5OH

Solution:1.2 x 10 ~2 mOr^HsOH

—vrt-erH^QH = 0.060LC2H5Or0.200 —

L solutionCheck: The units of the answer (L C2H5OH) are correct. The magnitude is reasonable for the amountand volume of solution.

4.57 Given: 400.0 mL; 1.1 M NaNO3 Find: g NaNO3

Conceptual Plan: mL solution -» L solution, then volume solution x M = mol NaNO3

on (L) x M = mol

then mol NaNO3 -» g NaNO3

85.01gNaNO3

mol NaNO3

1L l.lfnotNaNOa 85.01 gSolution: 400. Omt-sokrtienx—— — -x — — — — —\- , .Tr, = 37gNaNO31000 mi LToitrtiGnCheck: The units of the answer (g NaNO3) are correct. The magnitude is reasonable for the concentrationand volume of solution.


li> 0.102 moHSttOg 2irTDtNa3P04 1L 1000 mLSolution: 95.4 rrrfrGtiGl? x — — — — x - — - - x - x - x

1000 rrtL I K SrfTOhettOa O.l/SmDH^POi I K= 37.1mLNa3PO4

Check: The units of the answer (mL Na3PO4) are correct. The magnitude of the answer is reasonable sincethe concentration of Na3PO4 is greater.

4.64 Given: 125 mL, 0.150 M Co(NO3)2; 0.150 M Li2S Find: volume Li2SConceptual Plan: mL Co(NO3)2 -* L Co(NO3)2 -* mol Co(NO3)2 -> mol Li2S -> L Li2S — » mL Li2S

1 L 0.150 mol Co(NO3)2 1 mol Li2S 1 L 1000 mL1000 mL L lmolCo(NO3)2 0 .155 mol Li2S L

Solution:O.lSOmol-eofNQ^ 1 ffloH^S IK lOOOmL

= 125 mL Li2SCheck: The units of the answer (mL Li2S) are correct. The magnitude of the answer is reasonable since theconcentrations are the same and the mole ratio is 1:1.

!

4.65 Given: 25.0 g H2; 6.0 M H2SO4 Find: volume H2SO4

Conceptual Plan: g H2 -> mol H2 -> mol H2SO4 -» L H2SO4

2 .016 g H2 3 mol H2SO4 1L1 mol H2 3 mol H2 6 .0 mol H2SO4

Ifnrrf-H? Smot-HTSGU 1LSolution: 25.0 g-HjX- ~x~ -x- 7^7- = 2.1LH2SO42. - -Check: The units of the answer (L H2SO4) are correct. The magnitude is reasonable since there are approxi-mately 12 mol H2 and the mole ratio is 1:1.

4.66 Given: 25.0 g Zn, 275 mL solution Find: M ZnCl2Conceptual Plan: g Zn — * mol Zn — * mol ZnCl2 — > M ZnCl2

65 .41 g Zn 1 mol ZnCL, mol ZnCL,

1 mol Zn 1 mo' Zn volume solutionSolution:

1 frtdhZn 1 mol ZnCl225.0g-Znx-— —.x-— ——-- = 0.3822 mol ZnCl265.41g^Zn 1 ffTDi-Zn0. 3822 mol ZnCl2 1000 irA

275 M. T" :

Check: The units of the answer (M ZnCl2) are correct. The magnitude is reasonable because the stoichiometryis 1:1 and the mol Zn is less than 0.5.

Types of Aqueous Solutions and Solubility

I 4.67 ' i (a) CsCl is an ionic compound. An aqueous solution is an electrolyte solution, so it conducts electricity.

(b) CH3OH is a molecular compound that does not dissociate. An aqueous solution is a nonelectrolytesolution, so it does not conduct electricity.

(c) Ca(NO3)2 is an ionic compound. An aqueous solution is an electrolyte solution, so it conducts electricity.

(d) C^HÔs is a molecular compound that does not dissociate. An aqueous solution is a nonelectrolytesolution, so it does not conduct electricity.

4.68 (a) MgBr2 is an ionic compound. An aqueous solution is a strong electrolyte.

(b) C^HÔn is a molecular compound that does not dissociate. An aqueous solution is a nonelectrolyte.

(c) Na2CO3 is an ionic compound. An aqueous solution is a strong electrolyte.

(d) KOH is a strong base. An aqueous solution is a strong electrolyte.


4.69 (a) AgNO3 is soluble. Compounds containing NO3~ are always soluble with no exceptions. The ions inthe solution are Ag +(aq) and NO3~ (aq) .

(b) Pb(C2H3O2)2 is soluble. Compounds containing C2H3O2~ are always soluble with no exceptions. Theions in the solution are Pb 2+(aq) and C2H3O2~(ag).

(c) KNO3 is soluble. Compounds containing K+ are always soluble with no exceptions. The ions in solutionare K+(aq) and NO3~ (aq).

(d) (NH4)2S is soluble. Compounds containing NH4 + are always soluble with no exceptions. The ions in

solution are NH±+(aq) and S 2~~ (aq).

4.70 (a) Agl is insoluble. Compounds containing I~ are normally soluble but Ag+ is an exception.

(b) Cu3(PO4)2 is insoluble. Compounds containing PO43~ are normally insoluble and Cu2+ is not an exception.

(c) CoCO3 is insoluble. Compounds containing CO32~ are normally insoluble and Co2+ is not an exception.

(d) K3PO4 is soluble. Compounds containing PC^3" are normally insoluble, but K+ is an exception. Theions in solution are K+(aq) and PO4 *~ (aq).

Precipitation Reactions

(a) Lil(aq) + BaS(aq) — > Possible products: Li2S and BaI2. Li2S is soluble. Compounds containing S2~ arenormally insoluble but Li+ is an exception. BaI2 is soluble. Compounds containing I~ are normally sol-uble and Ba 2+ is not an exception. Lil(aq) + BaS(aq) —* No Reaction

(b) KCl(aq) + CaS(aq) —> Possible products: K2S and CaCl2. K2S is soluble. Compounds containing S2~are normally insoluble but K+ is an exception. CaCl2 is soluble. Compounds containing Cl~ are nor-mally soluble and Ca2+ is not an exception. KCl(aq) + CaS(aq) — > No Reaction

(c) CrBr2(aq) + Na2CO3(aq) -» Possible products: CrCO3 and NaBr. CrCO3 is insoluble. Compoundscontaining CO3

2~ are normally insoluble and Cr2+ is not an exception. NaBr is soluble. Compoundscontaining Br~ are normally soluble and Na+ is not an exception.CrBr2(flg) + Na2CO3(a^) -> CrCO3 (s) + 2 NaBr(^)

(d) NaOH(aq) + FeQ3(aq) -» Possible products NaCl and Fe(OH)3. NaCl is soluble. Compounds con-taining Na"1" are normally soluble, no exceptions. Fe(OH)3 is insoluble. Compounds containing OH~are normally insoluble and Fe3+ is not an exception.3 NaOHfag) + FeCl3(a<7) -» 3 NaCl(aq) + Fe(OH)3(s)

4.72 (a) NaNO3(a<7) + KCl(aq) -> Possible products: NaCl and KNO3. NaCl is soluble. Compounds contain-ing Na+ are always soluble, no exceptions. KNO3 is soluble. Compounds containing K+ are alwayssoluble, no exceptions. NaNO^(aq) + KCl(aq) — * No Reaction

(b) NaCl(aq) + Hg2(C2H3O2)2(fl<7) -> Possible products: NaC2H3O2 and Hg2Cl2. NaC2H3O2 is soluble.Compounds containing Na+ are always soluble, no exceptions. Hg2Cl2 is insoluble. Compounds con-taining Cr are normally soluble but Hg2

2+ is an exception.2 NaCl(fl«j) + Hg2(C2H302)2(aj) -* 2 NaC2H3O2(fl<7) + Hg2Cl2(s)

(c) (NH4)2SO4(fl<?) + SrC\2(aq) -» Possible products: NH4C1 and SrSO4. NH4C1 is soluble. Compoundscontaining NH4

+ are always soluble, no exceptions. SrSO4 is insoluble. Compounds containing SO^ 2~are normally soluble but Sr 2+ is an exception. (NH4)2SO4(fl(?) + SrCl2(a<7) -» 2 NH4Cl(fl(?) + SrSO4(s)

(d) NH4Cl(a^) + AgNO3(aq) -» Possible products: NH4NO3 and AgCl. NH4NO3 is soluble. Compoundscontaining NH4

+ are always soluble, no exceptions. AgCl is insoluble. Compounds containing Cl~ arenormally soluble, but Ag+ is an exception. NH4Cl(fl(y) + AgNO3(o^) -> NH4NO3(ag) + AgCl(s)

Chapter 4 Chemical Quantities and Aqueous Reactions

4.74

(a)

(b)

(c)

(d)

(a)

(b)

(c)

(d)

K2CO3(a<7) + Pb(NO3)2(aq) -» Possible products: KNO3 and PbCO3. KNO3 is soluble. Compoundscontaining K+ are always soluble, no exceptions. PbCO3 is insoluble. Compounds containing CO3

2~are normally insoluble and Pb2+ is not an exception.K2C03(fl<7) + Pb(N03)2(fl<7) -» 2 KN03(fl<7) + PbCO3(s)

Li2SO4(flg) + Pb(C2H3O2)2(a<7) -» Possible products: LiC2H3O2 and PbSO4. LiC2H3O2 is soluble.Compounds containing Li+ are always soluble, no exceptions. PbSO4 is insoluble. Compounds con-taining SO4

2~ are normally soluble but, Pb2+ is an exception.Li2SO4(a<7) + Pb(C2H3O2)2(fl</) -> 2 LiC2H3O2(fl<7) + PbSO4(s)

Cu(NO3)2(<z<7) + MgS(s) — > Possible products: CuS and Mg(NO3)2. CuS is insoluble. Compounds contain-ing S2~ are normally insoluble and Cu2+ is not an exception. Mg(NO3)2 is soluble. Compounds containingNO3" are always soluble, no exceptions. Cu(NO3)2(aq) + MgS(s) — » CuS(s) + Mg(NO3)2(fl<y)

Sr(NO3)2(a<7) + KI(ag) — * Possible products: SrI2 and KNO3. SrI2 is soluble. Compounds containing I~are normally soluble and Sr2"1" is not an exception. KNO3 is soluble. Compounds containing K+ arealways soluble, no exceptions. Sr(NO3)2(fl<7) + Kl(aq) — * No Reaction

NaCl(o<7) + Pb(C2H3O2)2(fl<7) -> Possible products NaC2H3O2 and PbCl2. NaC2H3O2 is soluble.Compounds containing Na+ are always soluble, no exceptions. PbCl2 is insoluble. Compounds con-taining Cl~ are normally soluble but Pb2+ is an exception.2 NaG(aq) + Pb(C2H3O2)2(fl<?) -> 2 NaC2H3O2(«?) + PbCl2(s)

K2SO4(fl^) + SrI2(a<7) — > Possible products: KI and SrSO4. KI is soluble. Compounds containing K+ arealways soluble, no exceptions. SrSO4 is insoluble. Compounds containing SO4

2~ are normally soluble,but Sr2+ is an exception. K2SO4(aq) + Srl2(aq) -» 2 KI(aq) and SrSO4(s)

CsCl(aq) + CaS(aq) — » Possible products: Cs2S and CaCl2. Cs2S is soluble. Compounds containing S2~are normally insoluble but Cs+ is an exception. CaCl2 is soluble. Compounds containing Cl~ are nor-mally soluble and Ca2+ is not an exception. CsCl(aq) + CaS(aq) — » No Reaction

Cr(NO3)3(a<7) + Na3PO4(fl(/) -* Possible products: CrPO4 and NaNO3. CrPO4 is insoluble.Compounds containing PQj3" are normally insoluble and Cr3+ is not an exception. NaNO3 is soluble.Compounds containing Na+ are always soluble, no exceptions.Cr(NO3)3(«?) + Na3PO4(fl<7) -> CrPO4(s) + 3

Ionic and Net Ionic Equations

4.75 (a)

(b)

(c)

(d)

H+(aq)H+(aq) + OH~(aq)

S2~ (aq)Cu+(aq) + S2' (aq)2+

H2O(/)

Cu2+(aq)» CuS(s)

OH' (aq) + H+(aq)H+(aq) + OH- (aq) + -» H2O(/)

6 Na+(fl<?) + 2 PO4 3

3Ni2+(fl<?) + 2PO4:

4.76 (a) 2 K+(aq) + SO4:

3 Niz+(aq) + 6 €•* Ni3(P04)2(s)

+ Ca2+(aq) + 2f (aq)-» CaS04(s)

(b)

(c)

+ Q- (aq) ++ OH' (aq)

Ag+(fl<?) + N03- (oj) + Na+(a<7) ++ a- (aj) -» AgCl(s)

H20(/) +1

2 Q- (fl<?) -» CuS(s) + Mg 2+(aq) + 2 &~ (aq)

H20(/) 4

"(a?) -» Ni3(P04)2(s

* CaSO4(s) + 2:

- H2O(/)

AgCl(s)

(aq)

+ 2

+ OH" (aq)H2O(/)

(aq)

(09)


(d) 2H+(aq) + aGaHaejj- (aq) + 3K+(aa) + CO32'(aq) -» H2O(/) + CO2(g) + 3K>a) + 3G&&- (aq)

* H2O(l) + CO2(g)

4.77 Hg22>a) + 2N03- (aq) + 3Nft>a) + 2 Cl~ (flfl) -* Hg2Cl2(s) + 3Ne>o) + 3W03-

Hg22>a) + 2 Cl- (aq) -» Hg2Cl2(s)

4.78 Fb2+(aq) + 3N©-f (017) + 2K+(aq) + SO4 2" (aa) -» PbSO4(s) + 2K+(ao) + 2N©3~ (aa)

Pb2+(«o) + SO4 2' (aq) -* PbSO4(s)

Acid-Base and Gas-Evolution Reactions

4.79 Skeletal reaction: HBr(aa) + KOH(aa) -» H2O(/) + KBr(aa)acid base water salt

Net ionic equation: H+(aa) + OH- (aq) -> H2O(/)

4.80 Skeletal reaction: HNO3(aa) + Ca(OH)2(aa) -^ H2O(/) + Ca(NO3)2(aa)acid base water salt

Balanced reaction: 2 HNO3(aa) + Ca(OH)2(ao) -> 2 H2O(/) + Ca(NO3)2(flo)Net ionic equation: H+(aa) + OH~ (aq) -» H2O(/)

(a) Skeletal reaction: H2SO4(ao) + Ca(OH)2(ao) -^ H2O(/) + CaSO4(s)acid base water salt

Balanced reaction: H2SO4(ai?) + Ca(OH)2(oo) -» 2 H2O(/) + CaSO4(s)

(b) Skeletal reaction: HClO4(ao) + KOH(aq) -» H2O(/) + KClO4(aa)acid base water salt

Balanced reaction: HClO4(fl0) + KOH(flfl) -> H2O(/) + KClO4(aa)

(c) Skeletal reaction: H2SO4(aq) + NaOH(aa) -* H2O(/) + Na2SO4(ao)acid base water salt

Balanced reaction: H2SO4(ao) + 2 NaOH(ao) -» 2 H2O(/) + Na2SO4(ao)

4.82 (a) Skeletal reaction: HI(aa) + LiOH(ao) -> H2O(/) + Lil(aa)acid base water salt

Balanced reaction: HI(aa) + LiOH(aa) -> H2O(/) + Lil(aa)

(b) Skeletal reaction: HC2H3O2(aa) + Ca(OH)2(aa) -* H2O(/) + Ca(C2H3O2)2(flo)acid base water salt

Balanced reaction: 2 HC2H3O2(flo) + Ca(OH)2(fla) -> 2 H2O(/) + Ca(C2H3O2)2(a(?)

(c) Skeletal reaction: HCl(aa) + Ba(OH)2(«fl) -^ H2O(/) + BaCl2(aa)acid base water salt

Balanced reaction: 2 HCl(«a) + Ba(OH)2(ao) -» 2 H2O(/) + BaCl2(aa)

4.83 Given: 22.62 mL, 0.2000 M NaOH solution; 25.00 mL HC1O4 solution Find: M HC1O4 solutionConceptual Plan: mL NaOH -> L NaOH -> mol NaOH -> mol HC1O4

1L 0.200 mol NaOH 1 mol HC1O4

1000 mL L NaOH 1 mol NaOH

mol HCIO^ volume HC1O4 solution — » M

1 mol HC1O4M =

L HC1O4 solution1 L 0 .2000 mot-NaOH 1 mol HC1O4

Solution: 22.62 mtrNaOH x ———- x - -x- ^~ = 0 .004524 mol HC1O41000 fri LNaOH 1 -0.004524 mol HC1O4 1000 fnt,

-x- ' = 0. 18096 MHC1O4 = 0.1810 M HC1O425.00 frtL HC1O4 1 L

Check: The units of the answer (M HC1O4) are correct. The magnitude of the answer is reasonable since itis less than the M of NaOH.


(c) C1O3~. The oxidation state of Cl = +5, and the oxidation state of O = - 2. The oxidation state of O is nor-mally - 2, and the oxidation state of Cl is deduced from the formula since the sum of the oxidation statesmust equal the charge of the ion. (Cl ox state) + 3(O ox state) = - 1; (Cl ox state) + 3(- 2) = - 1, so Cl = +5.

(d) C1O4~. The oxidation state of Cl = +7, and the oxidation state of O = - 2. The oxidation state of O is nor-mally - 2, and the oxidation state of Cl is deduced from the formula since the sum of the oxidation statesmust equal the charge of the ion. (Cl ox state) + 4(O ox state) = - 1; (Cl ox state) + 4(- 2) = - 1, so Cl = +7.

(a) 4 Li(s) + Qzfe) -» 2 Li20(s)Oxidation states; 0 0 + 1 - 2This is a redox reaction since Li increases in oxidation number (oxidation) and O decreases in num-ber (reduction). O2 is the oxidizing agent, and Li is the reducing agent.

(b) Mg(s) + Fe2+H) -» Mg2^) + Fe(s)Oxidation states; 0 + 2 + 2 0This is a redox reaction since Mg increases in oxidation number (oxidation) and Fe decreases in number(reduction). Fe2+ is the oxidizing agent, and Mg is the reducing agent.

(c) Pb(NO3)2(fl<7) + Na2SO4(fl<7) -» PbSO4(s) + 2Oxidation states; + 2 + 5 - 2 + 1 + 6 - 2 +2+6-2 +1 +5-2This is a not a redox reaction since none of the atoms undergoes a change in oxidation number.

(d) HBr(fl^) + KOH(fl^) -» H2O(/) + KBr(aq)Oxidation states;+l -1 +1-2+1 +1-2 +1 -2This is a not a redox reaction since none of the atoms undergoes a change in oxidation number.

4.92 (a) Al(s) + 3 Ag+(aq) — Al3+(aq) + 3 Ag(s)Oxidation states; 0 + 1 + 3 0This is a redox reaction since Al increases in oxidation number (oxidation) and Ag decreases in number(reduction). Ag+ is the oxidizing agent, and Al is the reducing agent.

(b) S03(g) + H20(/) - H2S04(a<?)Oxidation states; +6 -2+1-2 +1 +6 -2This is a not a redox reaction since none of the atoms undergoes a change in oxidation number.

(c) Ba(s) + Cl2(g) -* BaCl2(s)Oxidation states; 0 0 +2-1This is a redox reaction since Ba increases in oxidation number (oxidation) and Cl decreases in num-ber (reduction). C12 is the oxidizing agent, and Ba is the reducing agent.

(d) Mg(s) + Br2(/) -^ MgBr2(s)Oxidation states; 0 0 +2-1This is a redox reaction since Mg increases in oxidation number (oxidation) and Br decreases in number(reduction). Br2 is the oxidizing agent, and Mg is the reducing agent.

4.93 (a) Skeletal reaction: S(s) + O2(g) -» SO2(g)

Balanced reaction: S(s) + O2(g) -» SO2(g)

(b) Skeletal reaction: C3H6(g) + O2(g) -» CO2(g) + H2O(#)

Balance C: C3H<,(g) + O2(g) -» 3CO2(g) + H2O(g)

Balance H: C3H6(£) + O2(#) -» 3CO2(g) + 3H2O(g)

Balance O: C3H6(g) + 9/2 O2(g) -» 3CO2(g) + 3H2Ofe)

Clear fraction: 2C3H6fe) + 9O2(g) -» 6CO2(g) + 6H2O(g)

(c) Skeletal reaction: Ca(s) + O2(g) -* CaO(s)Balance O: Ca(s) + O2(g) -» 2CaO(s)

Balance Ca: 2Ca(s) + O2(g) -* 2CaO(s)

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