Ch4Answers - _ _ at}; f. inc:er .é-J'fik—dfl -____...

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Unformatted text preview: _ _ at}; f. inc:er .é-J'fik—dfl -____ AMC_:)_ T: f; " __3e_*_+-4I émgofiiwga r-r;_o___ [5} i........._.«..,...,..~. v. s,‘ wt. 4.... .0 a. , A mud-.4” Mum: iM'mflul -. w... wear... I“:- M h...“ .._ .- m. I 50 CHAPTER 4 SOLUTION STOICHIOMETRY “M 30. 31. 2 NaaP04(a<D + 3 FWQHM) " Pb3(PO4)z(s) + 6 N3N03(aQ) 0.250 mol Pb(NO3)2 -2mol Na3PO4 lLNa3PO4 X __ x ———___ = 0.250 L L 3 mol Pb(N03)2 0.100 mol Na3PO4 0.1500L X = 250. mL Na3P04 2 ASN03(3‘1) + CaClzth) —* 2 AgCl(s) + Ca(NOa)2(aq) 0.20 mol AgNO3 L 0.15 mol CaCl2 mol AgNO; = 0.1000 L x = 0.020 mol AgNO3 mol Cac12 = 0.1000 L x = 0.015 mol CaClz The required mol AgNO, to mol CaCl, ratio is 2:1 (from the balanced equation). The actual mol ratio present is 0.020/0.015 = 13 (13:1). Therefore, AgNO3 is the limiting reagent. ‘ 1 mol AgCl 143.4 g AgCl _ massA l=0.020molA o x ———_x————-2.9 A 1 g0 gN 3 'lmolAgNOB molAgCl 8 EC The net ionic equation is: Ag+(aq) + C1‘(aq) -> AgCl(s). The ions remaining in solution are the unreacted Cl' ions and the spectator ions, NO; and Ca2+ (all Ag” is used up in forming AgCl). The mol of each ion present initially (before reaction) can be easily determined from the mol of each reactant. 0.020 mol AgNO, dissolves to form 0.020 mol Ag” and 0.020 mol N031 0.015 mol CaCl2 dissolves to form 0.015 mol Ca2+ and 2(0.015) = 0.030 mol Cl". . I mol unreacted CI“ = 0.030 molCl' initially - 0.020 mol Cl' reacted = 0.010 mol Cl' unreacted 0.010 mol Cl ' MC] . = W = = 0050 MC]- total volume 0.1000L+0.1000L ‘ The molarity of the spectator ions are: 0020 “101 NO; _ 0.015 mol Ca 2* = 0.075 M Ca2+ = 0.10 MNo,-; Mcflz. M . = “03 0.20001. 0.2000L a. Cu(No,),(aq) + 2 KOH(aq) ~+ Cu(OH)2(s) + 2 KN03(aq) Solution A contains 2.00 L x 2.00 mol/L = 4.00 mol Cu(N03)2 and solution B contains . 2.00 L X 3.00 mol/L = 6.00 mol KOH. Lets assume in our picture that we have 4 formula units of Cu(NO3)2 (4 Cu2+ ions and 8 NO; ions) and 6 formula units of KOH (6 K+ ions and 6 OH' ions). With 4 Cu2+ ions and 6 OH' ions present, then OH' is limiting. One Cu“ ion remains as 3 Cu(0H)2(s) formula units form as precipitate. The following drawing summarizes the ions that remain in solution and the relative amount of precipitate that forms. Note that K+ and NO; ions are spectator ions. In the drawing, V1 is the volume of solution A or B and V2 is the volume of the combined solutions with V2 = 2‘V1. The drawing exaggerates the amount of precipitate that would actually form. CHAPTER 4 SOLUTION STOICHIOMETRY 51 33. 34. Cu(0H)2 Cu(OH)2 b. The spectator ion concentrations will be one-half of the original spectator ion concentrations in the individual beakers because the volume was doubled. Or using moles, MK . = MIS— 8.00 mol No; , , _4-00L = 1.50 M and MNOS- = W = 2.00 M The concentration of OH' ions Will be zero since OH' is the limiting reagent. From the drawing, the number of Cu” ions will decrease by a factor of four as the precipitate forms. Since the volume of solution doubled, the concentration of Cu2+ ions will decrease by a factor of eight after the two beakers are mixed: 1 M012. = 2.00 [ E) = 0.250 M Alternately, one could certainly use moles to solve for M012 : 3.00 mol OH ' x lmol Cu 2* L 2 mol OH ' 2.00 mol Cu 2* L excess Cu2+ present after reaction = 4.00 mol - 3.00 mol = 1.00 mol Cu2+ excess 1.00 mol Cu2+ - .= ——— = 0.250 M “2 2.00 L + 2.00 L mol Cu2+ reacted = 2.00 L X = 3.00 mol Cu2+ reacted mol Cu2+ present initially = 2.00 L X = 4.00 mol Cu2+ present initially MC 1 mol Cu(OH)2 97.57 g Cu(OI-I)2 — x _.__.____.___ =293 Cu 0 2molKOH mol Cu(0H)2 g ( H” massvof precipitate = 6.00 mol KOH x 0.200 mol Nazszo3 X lmol AgBr x 187.8 gAgBr L 2 mol NaZSZO3 mol AgBr 1.00LX =18.8gAgBr All the T1 in T11 came from T1 in T12804. The conversion from TlI to T12804 utilizes the molar masses of each compound. "Jim 2.- 1 wa-mm‘. CHAPTER4 SOLUTION STOICHIOMETRY , t . 53 Acid-Base Reactions 38. a. NH3(aq) + HNO3(aq) -> NI-LN03(aq) (molecular equation) NH3(aq) + I-P(aq) + NO3‘(aq) “r NH4”(aq) + N03‘(aq) (complete ionic equation) NH3(aq) + I-P(aq) —> NHJaq) (net ionic equation) b. Ba(OH)2(aq) + 2 HCl(aq) —> 2 H200) + BaClz(aq) Ba2+(aq) + 2 0H‘(aq) + 2 I-F(aq) + 2 Cl'(aq) —> Ba2+(aq) + 2 Cl'(aq) + 2 H200) ' OH'(aq) + maq) -> H200) c. 3 HClO,,(aq) + Fe(OH)3(s) -> ~3 H200) + Fe(ClO4)3(aq) 3 mm) +3 c10;-(aq) + Fe(OI-I),(s) —> 3 H200) + Fe3*(aq) + 3 ClO4'(aq) 3 H“ (m!) + Fe(0H)3(S) ‘* 3 H200) + F63+(8‘I) d. Ag0H(s>+HBr(aq>+AgBr(s)+Hzoa) mm + Hcaq) + Br-(aq) + AgBr(s) + Hzoa) Ag0H(s> + H’(aq) + Br-(acv e AgBr(s) + H200) 39. a. Perchloric acid reacted with potassium hydroxide is a possibility. HClO‘(aq) + KOH(aq) —> H,O(l) + KC104(aq) b. Nitric acid reacted with cesium hydroxide is a possibility. HNO;(aq) + CsOH(aq) —> H200) + CsN03(aq) c. Hydroiodic acid reacted with calcium hydroxide is a possibility. We get the empirical formula from the elemental analysis. Out of 100.00 g caiminic acid there are: 2HI(aq)+Ca(0H)2(aq) .—>2Hzo(1>+Ca12@ 53.66 g c x M = 4.468 mol c; 4.09 g H x M = 4.06 mol H 12.011 g c 1.008 gH 42.25 g o x w = 2.641 mol 0 15999 g o Dividing the moles by the smallest number gives: fl§§ =1,692; 3% =154 2.641 2.641 These numbers don’t give obvious mol ratios. determine the mol C to mol H ratio: 0 4.468 ___ 1.10: E . 4.06 10 c So let's try = 0.406 as a common factor: 4'468 = 11.0; = 10.0; = 6.50 10 ’ 0.406 0.406 0.406 Therefore, CMHZOOI3 is the empirical formula. We can get molar mass from the titration data. The balanced reaction is HA(aq) + 0H ‘(aq) i ' H20(l) + A'(aq) where HA is an abbreviation for carminic acid, an acid with one acidic H"’. 1802 x 10-3 L soln x 0.0406 mol NaOH x 1 mol carininic acid L soln mol NaOH = 7.32 x 10‘4 mol carminic acid E. Molamass: 0.3602g = 492g r: . 7.32x10'4mol . mol I The empirical formula mass of CnHzoOB. = 22(12) + 20(1) + 13(16) = 492 g. Therefore, the molecular formula of carminic acid is also CnHzoOu. ii 41. If we begin with 50.00 mL of 0.100 M NaOH, then: f; 50.00 ,x 10-3 L x W = 5.00 x 10-3 mol NaOH to be neutralized. a. NaOH(a0) + HC1(aq) —> NaCl(aq) + H200) ; 5.00 x 10-3 molNaOH x w {M = 5.00 x 10-2 L or 50.0 mL - 1 - mol NaOH 0.100 mol b. 2 NaOH(aq) + sto,(aq) » 2 H200) + Na,so,(aq) 1 mol so ‘ ' H 5.00 x 10-3 mol NaOH x —I_{2_3 x 1 L “1" = 2.50 x 10-2 L or 25.0 mL 2 mol NaOH 0.100 mol HZSO3 c. 3 NaOH(aq) + H3PO4.(aq) -* Na3P04(aq) + 3 H200) 1 m01H3PO4 x l soln 3 mol NaOH 0.200 mol H3PO4 5.00 X 10‘3 mol NaOH X = 8.33 X 10'3 L or 8.33 mL I V ' d. HNo,(aq) + NaOH(aq) —> H200) + NaNo,(aq) 1 mol HNO3 1 L soln x ——h = 3.33 ><10‘2 L or 33.3 mL 5.00 X 10'3 mol NaOH x . mol NaOH 0.150 mol HNO3 58 . CHAPTER4 SOLUTION STOICHIOMETRY b. Assign O a -2 oxidation state, which gives nickel a +4 oxidation state. Ni, +4; O,'-2. c. KJe(CN)6 is composed of K+ cations and Fe(CN)6" anions. Fe(CN)6“ is composed of iron and CN' anions. For an overall anion charge of -4, iron must have a +2 oxidafioa state d. (NI-I02HP04 is made of NH; cations and I-IPO42 ~2 as the oxidation state of O. In NH], x + In HPOE‘, +1 + y + 4(-2) = -2, y = +5 = ' anions. Assign +1 as oxidation state of H and 4(+1) = +1, x = -3 = oxidation state of N. oxidation state of P. ‘e. o, .2; P, +3 f. 0,—2; Fe,+8/3 g. 0,4; F, -1;- Xe,+6 h. F,I-1; s,+4 i. 0,-2; C, +2 j. ' Na, +1; 0,-2; C, +3 52. a. HBr:H,+l; Br,'-1 . b. HOBr: H, +1; O, -2; For Br, +1 +1(-2)+x= 0, x = +1 c. Brz: Br, 0 HBrO4: H, +1; 0, -2; For Br, +1 + 4(-2) + x = O, x = +7 'e. Bng: F, 4;, ForBr,x+3(-1)= o, x=+3. 53. 2a. .3 b. .3 c. 2(x)+4(+1)=0,x=-2 d. +2 e. +1 .f. +4 g. +3 ’ 11. +5' i. 0 a. UOK’: O, -2; For U, x + 2(-2) = +2, 1: = +6 |:, L A5203: 0, -2; For As, 2(x) + 3(-2) = 0, x = +3 c. NaBiOa: Na, +1; 0, -2; For Bi, +1 + x + 3(4) = o, a}: +5 d. A54: As, 0 e. HAsOZ: assignH=+1andO=-2; FotAs, +1 +x+2(—2)=0; x=+3 Mg2P207z' Composed of Mg2+ ions and P207“ ions. Oxidation states are: Mg, +2; 0, .2; P, +5 3. N828203: Composed of Na" ions and 8,032" ions. Na, +1; 0, -2; S, +2 CHAPTER 4 SOLUTION STOICHIOMETRY ' ‘ . . 59 M 55. 56. h. HgZCIz: Hg, +1; Cl,-1 i. Ca(NO3)2: Composed of Ca2+ ions and N03“ ions. Ca, +2; 0, -2; N, +5 To determine if the reaction is an oxidation-reduction reaction, assign oxidation states. If the oxidation states change for some elements, then the reaction is a redox reaction. If the oxidation states do not change, then the reaction is not a redox reaction. In redox reactions the species V oxidized (called the reducing agent) shows an increase in the oxidation states and the species reduced (called the oxidizing agent) shows a decrease in oxidation states. Redox? Oxidizing Reducing Substance Substance _ Agent Agent Oxidized Reduced a. Yes 02 ’ CH4 CH4 (C) 02 (O) b. Yes HCl ' Zn Zn HCl (H) c. No _ - - - - d. Yes 03 -NO NO (N) 03 (O) ' e. Yes H202 H202 H202 (0) H202 (O) f. Yes ‘ CuCl CuCl' ’CuCl (Cu) CuCl (Cu) g. No ’ - - - ' - h. No - - - - i. Yes SiCl4 Mg Mg - SiCl4 (Si) In c, g, and h no oxidation states change from reactants to products. a. The first step is to assign oxidation states to all atoms (see numbers above the atoms). -3 +1 0 +4 -2 +1 -2 C2H6 + 02 —> (:02 + H20 Each carbon atom changes from -3 to +4, an increase of seven. Each oxygen atom changes from, 0 to -2, a decrease of 2. We need 7/2 O-atoms for every Cgatom. CZHG + 7/2 02 -> CO2 + H20 Balancing the remainder of the equation by inspection: ” Csz(g) f 7/2 02(g) —* 2 C02(g) + .3 H20(g) or 2 C2Hs(g) + 7 02(3) _’ 4 C02(g) + 6 H20(g) - b. The oxidation state of magnesium changes fiom 0 to +2, an increase of 2. The oxidation state i of hydrogen changes from +1 to 0, a decrease of 1. We need 2 H-atoms for every Mg-atom. The balanced equation is: ' Mg(s) + ZHCKaQ) “* Mg2+(aCI) + 2 C1'(aq) + H2(g) 72 ' ——————_—__——______________ 77. 78. CHAPTER 4 SOLUTION STOICHIOMETRY We have equal masses of AgN O3 and KZCrO... Since the molar mass of AgN O3 is less than that of K2Cr04, then we have more mol of AgN 03 present. However, we will not have twice the mol of AgNO3 present as compared to KZCrO4 as required by the balanced reaction; this is because the molar mass of AgNO3 is no where near one-half the molar mass of K2Cr04. Therefore, AgNO3 is limiting. ‘ 2molAgNO3 x 169.9g ‘ mass AgNO3 = 1.000 mol AgZCrO4 X = 339.8 g AgNO3 mol AgZCrO4 moliaigNo3 Since equal masses of reactants are present, then 339.8 g K2Cr04 were present initially. 1 mol KZCI'O4 x 2 mol K + 33” “2904 x 194 20 fires“ mol K + . . - g m0 r MK. = ———__ = 2 4 =7.000MK+ total volume 1 0.5000 L 1 molK7C1'O4 1 mol Crof' ‘ b. mol Cr042' = 339.8 gKZCrO4 x - x ———_ = 1.750 mol CrOf' present initially < ' 19420 g “ml K2Cr04 1' mol Cr042' Cf042' in = AgZCI‘O4 x —‘_—‘ 1 mol Ag2CrO4 = 1.000 mol Cr042' precipitate _ _ excess mol CrO 2" _ MOO? = 4 = 1.750 mol 1.000 mol = 0.750 mol = 0.750M total volume 0.5000 L + 0.5000 L 1.0000 L Cr(N03)3(aCI) + 3 Na0H(aQ) —* Cr(0H)3(S) + 3 NaN 0301‘!) 1 mol Cr OH mol NaOH used = 2.06 g Cr(OH)3 x —Lfi x M; 6.00 x 10-2 molNaOH to form precipitate 103.02 g 1 mol Cr(OH)3 NaOH(aq) + HCl(aq) —> NaCl(aq) + H200) mol NaOH used = 0.1000 L x Min—(“lg x H‘flO—IL 4.00 x 10-2 mol NaOH to react with HCl L , mol HCI C -2 _2 MM)“: w = 6.00 x 10 me] + 4.00 x 10 mol ____ 2.00 MNaOH volume 0.0500 L First we will calculate the molarity of NaCl while ignoring the uncertainty. -3 —2 0.150 g x lmol = 2.57 x 10_3 moles; Molarity = 2.57 x 10 mol 2 2.57 x 10 mol NaCl 58.44 g V 0.1000L L 0.153 g x “ml , 58.44 g = 2.63 ><,10~~ mol NaCl 0.0995 L ‘ . L The maximum value for the molarity is: CHAPTER 4 SOLUTION STOICHIOMETRY v 73 ____________________.____————-—-—————————— 79. 80. 81. 1 mol ' 0'147gx58 4 250 10‘2 IN C1 The minimum value for the molarity is: = w 0.1005 L L The range of the molarity- is 0.0250 M to 0.0263 M or we can express this range as 0.0257 :1: 0.0007 M. The amount ofKI-[P used = 0.4016 g x 1 “‘01 = 1.967 x 10-3 mol KHP . . ’ 204.22 g v Since one mole of NaOH reacts completely with one mole of KHP, then the NaOH solution contains 1.967 x 10'3 mol NaOH. .. . _3 . _2 Molarity ofNaoH = W = W 25.06 x10’3L _ L . . _ 1.967 x 10'3m01 _ 7.865 x 10'21nolNaOH " Max1mum molarlty — ——-——-— — ———————— 25.01x10'3L L . . . _ 1.967x10'3mol _ 7.834x10'2molNaOH Minimum molarlty — —-—————————— — —-—————————— 25.11x10'3L L We can express this as 0.07849 :|: 0.00016 M. An alternate way is to express the molarity as 0.0785 :1: 0.0002 M. This second way shows the actual number of significant figures in the molarity. The advantage of the first method is that it shows that we made all of our individual measurements to four significant figures. Desired uncertainty is 1% of 0.02 or :1: 0.0002. So we want the solution to be 0.0200 :1: 0.0002 M or the concentration should be between 0.0198 and 0.0202 M. We should use a l-L volumetric flask to make the solution, They are good-to :l: 0.1%. We want to weigh out between 0.0198 mol and 0.0202 mol of K103. Molar mass ofKIO3 = 39.10 + 126.9 + 3(16.00) = 214.0 g/mol 215”) g = 4.237 g; 0.0202 mol x 214") g = 4.323 g (carrying extra sig. figs.) mol mol 0.0198 mol X We should weigh out between 4.24 and 4.32 g of K103. We should weigh it to the nearest mg or 0.1 mg. Dissolve the KIO3 in water and dilute to the mark in a One liter volumetric flask. This will produce a solution whose concentration is within the limits and is known to at least the fourth decimal place. 1 mol C6H807 molC 0 =0.250 CH0 x ——————— ,‘H” 7 gr“ 8 7 192.1gC6H807 = 1.30 x 10'3 mol caugo7 Let HxA represent citric acid where x is the number of‘ acidic hydrogens. The balanced neutralization reaction is: H..A(aq) + X 0H‘(aQ) "* X H200) + A"(aq) CHAPTER 4 SOLUTION STOICHIOMETRY 75 84. 275.0 mL X 0.300 M = 82.5 mmol H+; Let y = mL delivered by Y1 and z = mL delivered by y(0.150 M) + z(0.250 M) = 82.5 mmol OH' mmol H+ Note that units of molarity are mol/L but are also equal to mmol/mL. 275.0mL+y+z=655 mL, y+z=380. , z=380.-y y(0.150) + (380. - y)(0.250) = 82.5, y = 125 mL, 2 = 255 mL flow rates: Y a ~125—mL— = 2.06 mL/min and z —> 33313 = 4.20 mL/min 60.65 min . 60.65 min 85. Molar masses: KCl, 39.10 + 35.45 = 74.55 g/mol; KBr, 39.10 + 79.90 = 119.00 g/mol AgCl, 107.9 + 35.45 =_ 143.4 g/mol; AgBr, 107.9 + 79.90 = 187.8 g/mol Let x'= number of moles of KCl in mixture and y = number of moles of KBr in mixture. Since Ag+ + C1' -+ AgCl and Ag* + Br" —> AgBr, then x = moles AgCl and y = moles AgBr. Setting up simultaneous equations fiom the given information: ’ 0.1024 g= 74.55 x+ 119.0 y and 0.1889 g =’ 143.4 x+ 187.8 y Multiply the first equation by 1813?, then subtract from the second. 119.0 0.1889 = 143.4 x + 187.8 y -0.1616 = -1177 x -187.8 y 0.0273 = 25.7 x, x = 1.06 x 103 mol KCl 0 74.55 g KCl mol KCl 1.06 X 10'3 mol KCl X = 0.0790 g KCl Mass % KC1= 00790 g x 100 = 77.1%, % KBr = 100.0 — 77.1 = 22.9% 0.1024 g r 86. Pb2+ + 2 c1- —> 1>bc12 . f» ‘ 3.407 g ’ 1 mol PbCl 2+ .~ ' 3.407 g 1>bc12 x ————2 x 332131)— = 0.01225 mol Pb” 1"; ' 278.1 g PbCl2 1 mol PbCl2 ‘ w = 6.13 M Pb2+ (evaporated concentration) 2.00 x 10'3 L original concentration = W = 4.90 M 0.100 L CHAPTER 4 SOLUTION STOICHIOMETRY 87. a. CvlemmCl, + n Ag’r -> n AgCl; molar mass (AgC1)= 143.4 g/mol molar mass (PCB) = 1202.01) +(10—n) (1.008) + n(35.45) = 154.20 + 34.44 11 Since 11 mol AgCl are produced for every 1 mol PCB reacted, then n(143 .4) grams of AgCl will be produced for every (154.20 + 34.44 n) grams of PCB reacted. ‘ 212%; = _.__——154 21813;? 44 n or massMCI (154.20 + 34.44 11) = massPCB (143.4 11) b. 0.4971 (154.20 + 34.44 11) = 0.1947 (143.4 11), 76.65 + 17.12 11 = 27.92 11 76.65 = 10.80 n, n = 7.097 88. moles Cuso. = 87.6 mL x J— x 0500M = 0.0439 mol 1000mL 1L moles Fe e 2.00 g x 1 “101126 = 0.0358 mol 55.85 g The two possible reactions are: I. CuSO4(aq) + Fe(s) —> Cu(s) + FeSO4(aq) 11. 3 CuSO4(aq) + 2 Fe(s) —> 3 Cu(s) + Fe2(SO4)3(aq) Ifreaction I occurs, Fe is limiting and we can produce: 19219: x 92m =2.2s gen lmolFe lmol Cu If reaction II occurs, CuSO4 is limiting and we can produce: 0.0358 mol Fe ‘x 3 mol Cu X 63.55 g Cu = 2.79 g Cu 0.0439 mol CuSO4 X —————— 3:m01 CuSO4 1 mol Cu Assuming 100% yield, reaction I occurs since it fits the data best. 89. a. Flow rate = 5.00 X 104 US +3.50 X 103 L/s = 5.35 X 104 US b. CHC1 =3 M: 4.25 ppm HC1 ‘ . 5.35 x 104 c. 1 ppm = 1 mg/kg H20 = 1 mg/L (assuming density = 1.00 g/mL) . 4 X 60 min 60' X 1.80 x 10 L x 4.25 mg HC1 x 1g __ 2.20 x 105 gHCl mm s L 1000 mg . , 1mol HC1 x 1 mol CaO x 56.08g CaO = 1.69 X 106 g Cao 2.20 X 106 g HC1 X 36.46 g HC1 2 mol HC1 mol CaO ...
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This note was uploaded on 04/07/2008 for the course CHE 141 taught by Professor Kerber during the Fall '07 term at SUNY Stony Brook.

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Ch4Answers - _ _ at}; f. inc:er .é-J'fik—dfl -____...

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