MOOREHW4 - Chapter 4: Quantities of Reactauts and Products...

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Unformatted text preview: Chapter 4: Quantities of Reactauts and Products 12. Ausu-et'.‘ (a) 1.00 g (b) 2 for Mg. 1 for 03. and 2 for )IgO (c) 50 atoms ofMg Sontag and animation: Git-‘en the balanced equation for a reaction. identify the stoichiometric coefficients in this equation. and relate the quantity of products to reactants and Vice 1uersa. (a) The law of conservation of mass says that the total mass of the reactants is the same as the total mass of the products. (b) The stoichiometric coefficients are the numbers in front of each formula in the equation. (c) The stoichiometric coefficients can be interpreted as the related number of reactants. Use the formula stoichiometry' of 0:, to find the number of molecules that reacted. Use equation stoichiomeny to find out how many atoms of Mg reacted with that many 02 molecules. First. balance the equation: 2 Mg(s} + 03(g} — 2 Mg0(g) (a) The total mass of product. MgO. is l.00 g. The total mass of the reactants if mass of Mg plus mass of 0.2) that reacted must also be 1.00 g. due to the conservation of mass. (0} The stoichiomenic coefficients are 2 for Mg. 1 for 02. and 2 for MgO. (c) 50 atoms of oxygen make up 35 molecules of 01. since there are two 0 atoms in each molecule. Since the stoichiometiy is 1:1. that means 2'0 atoms of Mg were needed to react with this much 0?. J Rmsonobfe Astsu‘et' Check: Though it is an tuinecessaiy calculation. we can check the answer to part (a) by: calculating the masses of Mg and Ct2 that reacted to make l.00 gram of MgO, 1.00 g Mng 1 mol MgO x 1 mol Mg N 24.30210 g Mg 40.3044 g MgO 1 mol MgO 1 mol Mg — 0.503 g Mg 1 mol MgO 1 mol 0: 31.9988 2 01 _ ‘— x ' x ‘— ' — 0.39: 9 01 40.3044 g MgO 2mol MgO 1 mol 0; “ ' 1.00 g MgO >< Adding them up reproduces the product mass: 0.603 g — 0.39? g = 1.000 g. 34. Anni-er: (a) HINC21(aq)+ 2 NH3(g] —- NH4Cl(aq) + N2H4(aq) (b) (CH3):N1H3(£} + 2 312043) 3 N:(g) — 4 H20(g) — 2 (Dig) (c) Catizts) + l H:O(£} —-r Ca(0H)1{s} + C21H3(g] Shortage and Eapfonofimt.‘ Follow the method described in the solution for Question 30. (a) 2 HQNCHaq) + ‘3 NH3(g} —- 2NH4CTl(aq) + ‘3 N3H4(aq} Select order: Cl. N. H HENCllfiaq) + 2 NH3(g) —r- NH4Cl(aq] + ‘2 N3H4(aq} 1 C1 HENCllfiaq) + 2 NH3(g) —r- NH4Clfiaq) + N2H4fiaq) 3 N and a H (b) 2 (cngzNszm + 2 N304(g) —- 2 Nltg} + 2 H2043“) + 2 (03(2) Select order: H. c. 0.31 [CH3)2N3H3(F') + 3 N204fig‘} —I- 2N2(g) + 4 HEOIng} + '3 C02(g) 8 H (CHfllNgHEU) + 2 N204fig} —- 2202th + 4 H30(g} + 2 (:02th 2 c (CHSYJENEHEU') + 2 3130413) —-— ?N2(g) + 4 H30(g} + 2 cogtg) a o [CH3j2N3H3(I') + 2 N204(g} —-r 3 313th + 4 H30(g} + 2 cogtg) 631 (c) 2 CaCEIfis + 2 H3011?) —r- ‘3 Ca(OH_1_3(s) — '3 C2H3(g) Select order: Ca. C. O. H Cast) -- ‘3 HECIU) —I-— {MOI-Dis) + ‘3 CEHflg) 1 Ca Cacgts) + 2 H300) —- Ca(OH}:(s) + C:H_«.(g} 2 c: Cast) -- 2 H300) —I- CaIIOI-Dgts) + Cngfig} 2 O and4 H v’ Rertsouobfedtmt-‘er Check: (a) 8 H. 3 IN" 8: 1 C103) .3 C. 8 H.631 & S O (c) 1 Ca. 2 (.14 H 8: 2 O 58. Answer: (a) 699g (b) 526g Strategy and Eiipt‘oriarfon.‘ Given a balanced chemical equation and the mass of a reactant in kilograms. deternune the mass of one of the products produced and the mass of another reactant required. Use metric conversion factors to convert kilograms to grams. Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the product produced and the other reactant required. Then use molar masses to find the grams. 1000 g Felt); 1 mol Fegog x = 6261110113910; 1 kg rego3 159.688 g r9303 — ' 1:310}; x (a) The balanced equation says: 1 mol F9203 produces 1 mol Fe. 2 mol Fe 55.845 g Te —x = 699 g re 1 mol Fe303 1 mol Fe “ 6.26 mol l-"e303 x (b) The balanced equation says: 1 mol Felt}; requires 3 mol CO. ‘ l CO 28.0101 CO 6.2611101F6303XL x—g =5 6 Q ('0 1 mol R303 1 mol CO “ |\J v’ Reosonobfa Answer Check: The mass of iron produced is less than the mass of irot1(III'_J oxide it was produced floor. The mass of CO used is less than the mass of ironflll) oxide even with a larger stoichiometric coefficient because (‘0 contains lighter—weight atoms. 60. Answer: BaC]:.1.12081g E‘aSO4 .‘Sn'atagv and Eiipfonart'on.‘ Given a balanced chemical equation and the masses of both reactants. determine the limiting reactant and the mass of the product produced. Here. are use a slight variation of what the text calls "the mole method." We will calculate a directly comparable quantity. the moles of the desired product. Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. Then. use the moles of product produced from the limiting reactant and the molar mass of the product to find the grains. Exactly 1 gram of each reactant is present initially. The balanced equation says: 1 mol NaJSD; produces 1 mol BaSOq. 1 mol Na3304 lmol BaSCq 0 00104016 111013180 = I .. L. 4 1g Na1504 x _ >< _ ' 142.042 g haJSO4 lmol 3.9.3504 The balanced equation says: 1 mol BaC'l2 produces 1 mol BaSO4. 1 2 Each x 1 mol BaCl; lmol BaSO4 M —_>< = 000480231 mol BaSO4 ‘ 208.233 g BaLZl: 1 mol BaC'l: The number of BaSuD4 moles produced fiom BaL'Zl2 is smaller (000480231 mol c2: 000204016 mol). so Ba'C‘lJ is the limiting reactant. Find the mass of 000480231 mol BaSO4: 233.399 g Baso4 = 1.12081 2 1321504 1 mol BaSO4 ‘ 000480231 mol BaSO_— x v“ Reasonable Answer Check: There are fewer moles of BaSO'4 present than BaCl} and the equation needs the same amount of Bafsuo4 as BaC‘ll. so it makes sense that BatC'lE is the limiting reactant. . Answer: KDH, KOH .S‘nomgr and Eipionatfoarr Given equal moles of all the reactants. determine the limiting reactant. Given equal masses of all the reactants. determine the limiting reactant. When the same moles of ever}: reactant is present. the reactant with the largest stoichiometric coefficient is going to be the limiting reactant. In the second part. use the molar mass of each reactant to find their moles. then use the stoichiometm of the reaction to determine the moles ofa product formed from each reactant. The reactant that produces the least amount of product is the limiting reactant. First question: 5 mol of each reactant is present. The stoichionietric coefficients are: 4 for KOl-l. 2 for MnOl. l for O3 and 1 for Cl} The reactant with the largest stoichiometric coefficient is KOl-I. So. KOH is the limiting reactant. Second question: 5 grants of each reactant is present. Determine the moles of KCl the}; each form: 1 11101 KOH 2 mol KCl — x — 56.1056 g KOH 4 mol KOH 5 g Mnow x 1 mol Moo: 2 inol KC'l — X — ' 86.93.63 g 3-1110: 2 mol MnO: 1 mol 0: 3 niol KC'l x—x—= 0.3 inol KC'l 31.9988 g 03 l inol 03 _ 1 mol (21-. 2 mol KC'l :3 g C11 x—‘x— ' 10.906 g ('13 1 mol C13 5 g KOHx = {1.04 niol KC‘l = 0.06 mol KCl 5 g 0: = 0.1 niol KC‘l Only 0.04 mol of HCl is produced from KOH. so the KOH is the limiting reactant here._ too. v’ Reasonable Answer Check: The first question is easy. The reaction says that more ntoles ofKO1-1 are needed that an}! other reactant. so when equal moles of reactants are present. it runs out first. The differences in molar mass are insufficient to keep the large stoichiometric coefficient for KOH from making it the limiting reactant when equal masses of the reactants are present. 101. Anni-w: Ag"! Cu“. and 3'03— .Sfl‘ttt'egt: and Explanation: Given the moles of all the reactants. determine the limiting reactant and the ions present at the end of the leaction. Use the stoichiotnetry of the equation to determine the moles of a product formed from each reactant. The reactant that produces the least amount of product is the limiting reactant. The excess reactant and the products aie present at the end of the reaction. From that information. determine what ions are in the solution after the reaction is complete. 1 mol (110103)} 1 mol Cu J1 1.5 mol Cle =1. . inol CZuLNOfl; 1 mol C11[ND3}3 4.0 mol AgNO; >< ' 2 mol AgNO3 = 2.0 mol CutiNOfi] Only 1.5 mol CiltNOsj2 can be formed. so the limiting reactant is Cu and the excess reactant is AgNOS. That means the solution contains: Ag‘ ions. Cu:— ions. and N03: ions after the reaction is over. [The question only asked what ions are present not how many.) If we want to know how many. we need to determine the males of .i'igI‘JO3 actually ieacted: 3 mol AgNO; 1 mol C11 1.5 mol (11 X = 3.0 mol Apt-NOS reacted The mol ongN—Oj left = 4.0 mol AgNO3 initial — 3.0 AgNO3 mol reacted = 1.0 mol nigh—O3 left Quantitativeijt. the solution has 1.0 mol Ag" ions. 1.5 mol Cu3+ ions. and (1.0 mol + 2 x 1.5 mol 2} 4.0 mol N03‘ ions. r’ Reason-obit? Answer Check: The excess reactant is an ionic compound and one of the products is an ionic compound, so some of the ions present at the beginning are still present when the Cu runs out. The quantities ong' ions have dropped. since some of the silver atoms are now part of the Ag solid product. The quantity of NO; ions doesn't change. since neither N nor 0 are part of the solid product formed. 112. Annrer.‘ (a) tC‘H‘l (Mlflflg (c) Te'flflg .S‘fi'otagr and Explanation: Given a balanced chemical equation and the masses of both reactants. determine the limiting reactant. the mass of the product produced. and the mass remaining of the excess reactant when the reaction is complete. For part (a): Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. For part (b): Use the moles of product produced fi'om the limiting reactant and the molar mass of the product to find the grams. For part (c): From the quantity of limiting reactant. determine the moles of the other reactant needed for complete reaction. Convert that number to grams using the molar mass. Then subtract the quantity used fi'om the initial mass given to get the mass of excess reactant. Notice that 500 grams has 1 significant figure and 1300 grams has 2 significant figures. (a) The balanced equation says: 1 mol CH4 produces 3. mol H3. 500 g cm x—l “01 L'H4 >< 160—23 g CH4 3 mol H3 — = 90 mol Ha I mol CH4 ' The balanced equation says: 1 mol H30 produces 3. 111011-13. 1 1 H10 3 1 H1 “0—- mo - = 230 mol H3 1300 g Box 3:— - 18.0 5: g H30 111101 H30 The number of H2 moles produced fi'om C' -I_1 is smaller (90 mol -:::2 220 mol). so CH4 is the limiting reactant and H30 is the excess reactant. r} _ 'l (b) Find the mass Hi: 90 mol 1 xfl = 200 g H1, ' ' 1 mol H3 ' ' 1 s“- (C) 500gCH4X lmolCI—I4 leol HgOX18.GI__gHJO =600gH10 16.0433 g C' 4 1 mol CH4 1 mol H30 ‘ 1300 g H30 initial — 600 g H10 used up = .700 g H30 remains um‘eacted v“ Reosonobfe Answer Check: The small number of significant figures makes the uncertainty in this calculation somewhat high. However. the general expectations are still met. There is a larger mass ot‘l—LO present. and the molar masses of the reactants are about the same size. Since the equation indicates that equal moles of H30 and CH4 react. it makes sense that ('1-14 is the limiting reactant. In addition. the calculation in (c) proves that the initial mass of H30 was larger than required to react with all the C‘H4. C'hapte: 5: Cliemic a1 Reac:io:‘.s 34. Answer: (a) base, strong, K" and DH" (1)) base, strong, 31g} and OH" (c) acid, weak, H" and CID" ((1) acid. strong, H" and BI" (e) base, strong, Li+ and OH‘ (1‘) acid, weak; H"._ HSO3‘. and 8033‘ .S‘n'otagi' cmo‘ Explanation: Given some chemical formulas. identify if they are acids or bases identify whether they are weak or strong. and determine what ions produce with they dissolve in water. Acids produce H" in aqueous solution. Bases produce OH‘ in aqueous solution. In several of these cases. Table 5.1 on page 165 or Table 5.2 on page ITS can be used to determine whether a large or small amount of ions are produced. by determining if the compound is soluble and-"or weak or strong. In some cases. however. it is not possible to look up that information in Chapter 5. However. it is almost always true that. if an acid is not one of the common strong acids listed in Table 5.2. it is a weak acid. That is the case with the weak acids in this Question. If the acid or base is strong. it will ionize. If the acid or base is weak it will not ionize to a great extent remaining primarily in the molecular form. (a) ROI-I is a strong base. (Given in Table 5.2). producing K" and OH" ions when dissolved in water. (b) MgCCII-Ij2 is an insoluble ionic compound (Table 5.]. Rule 10) so few ions are produced in water. though the .‘v-IgEOPUJ that dissolves does ionize completely. So. practically. it may be considered weak. because the OH‘ ion concentration will never be very large. Technically. it may be considered to be strong. because all the dissolved .‘v-{gfiOHjIJ is ionized. producing Mg} and OH‘ ions when dissolved in water. (c) I—IC'lO is a weak acid. (It's not listed as a common strong acid in Table 5.2.}. It will not ionize very much. remaining mostly in the HClOan'} form. The small amount that does ionize will form H" and [‘10— ions. (d) HBr is a strong acid (Given in Table 5.2}. producing H" and Br" ions when dissolved in water. (e) LiOH is a strong base (Given in Table 5.2).. producing Li" and OH" ions when dissolved in water. \ I. It will not ionize very much. (f) H1503 is a weak acid (It's not listed as a common strong acid in Table 5.2. remaining mostly in the HfiOfiaq) form. The small amount that does ionize will form H". H303". and 8033‘ ions. v“ Rectsoaobia Answer Check: All of the ions produced are common ions. most of them are found in Figure 3.2 and Table 3.2. Most of these are given specifically in Table 5.2. If you want to look far into your chenustiy future. Table 16.2 on page TBS can also be used to confirm that the two weal-t acids given here in [ch and (D are indeed weak. 40. Answer: (a) Ox. # = —2, Ox. # S = +6 (1)) OX. # = —2, Ox. # H = +1.01 ti .V = +5 (c) Ox. :4 K = +1. Ox. : O=—1.0x.f.-' Mo = +'.-' (d) 03. a O = —2,0x. #H=+l (e) Ox. a Li = —1.0x..‘¢ O=—2. Ox. .s'H=—1 (f) Ox. is' C] [5—1. Ox. :1 H=-1._ Ox. .tC = 0 Sit'aragv and Exp-Taxation: Given several formulas of compounds. deteimine the oxidation numbers of the atoms in each of thent. For this Question. the rules spelled out in Section 5.4 on page 185 are used extensively. Several elements in compounds have predictable oxidation numbers (Rules 1 to 3'}. The oxidation numbers of the remaining atom(s) can be determined using the sum rule {Rule 4). The term "oxidation number“ is abbreviated below as "Ox. (a) 803 contains oxygen. Rule 3 gis'es us Ox. #5' O = —_‘. This is a molecule. so the sum of the oxidation numbers is zero. O=l><fOx.#S)-3><(—lj Therefore. Ox. # S = —6. (b) [-TIN'O3 contains oxygen and hydrogen. Rule 3 gives us Ox. # O = —2 and Ox. # H = —l. We use the sum rule to find the Ox. if N This is a molecule. so the sum of the oxidation numbers is zero. 0=1xI:—1'_J+l><(0x.#N)+3 ><(—3] Therefore. Ox. 7—? N = -5. (c) Kits-11104 contains a monatomic cation. K". and a polyatomic anion. Iv-InOf. Rule 2 gives us Ox. is K = —l. Rule 3 gives us Ox. if O = —3. We use the sum rule to find the Ox. if I‘vin. For a polyatonuc anion. the sum of the oxidation numbers is equal to the ions charge of 1—. —1 = 1 x (Ox. is Mn) - 4 x (—2) Therefore. Ox. 7—? Mn 2 —T. (d) I—IJO contains oxygen and hydrogen. Rule 3 gives us Ox. if O = —2 and Ox. e‘ H = —1. (e) LiOH contains a monatomic cation. Li'. and a diatomic anion. OI—I‘. Rule 2 gives us Ox. f—r' Li = —1. Rule 3 gives us Ox. .'=.' O = —2 and Ox. .'=.' H = —1. (ft C'l-IJCI2 contains carbon. hydrogen. and chlorine. Since chlorine makes a —1 ion when part of ionic compounds. we will assume its oxidation number is —1 (as was described in Section 5.4 on page 184 for the compound PC'ls'}. Rule 3 gives us Ox. if H = —l. We use the sum rule to find the Ox. 9': C. This is a molecule. so the sum of the oxidation numbers is zero. 0 = (Ox. if Cit—2 X (—1) — 3 x (—1) Therefore. Ox. 7—? C' = 0. v“ Reasooabi’e Answer Check: The non-metal elements farther to the right on the periodic table have consistently more negative oxidation numbers. The metallic elements and nonmetals farther to the left on the periodic table have more positive oxidation numbers. +3. Answer: Dnly reaction (b) is an oxidation—reduction reaction; oxidation numbers change. Reaction (a) is a precipitation reaction; reaction (c) is an acid—lJase neutralization reaction. Strategr and Efipioimtr'on: Given several reactions. determine if they are oxidation—reduction reactions and classify the remaining reactions. To decide if a reaction is an oxidation-reduction reaction. we need to see if any of the elements change oxidation state. Oxidation-reduction reactions are ones in which the atoms have different oxidation states before and after the reaction. If no change in oxidation state is observed. then it is not an oxidation—reduction reaction. (a) The ionic compounds representing reactants and products in this reaction all contain Cd}. C1". Na‘. and 83‘ ions. Therefore. this is not an oxidation—reduction reaction. The formation of insoluble C'dS classifies this reaction as a precipitation reaction. (b) The elements Ca and 03. both with zero oxidation states. are combined into an ionic compound. C'aO. with Ca:— and 03‘ ions. That means the oxidation numbers did change and this is an oxidation—reduction reaction. (c) The ionic compounds representing reactants contain Ca}. OH‘. H‘. and C'l‘ ions. The product ionic compound contains Ca:— and (‘1‘. The other product. water. has 0 in the —2 oxidation state and 1-1 in the —1 oxidation state. Therefore. this is not an oxidation—reduction reaction. The formation of water from the reaction of OH‘ and H‘ classifies this reaction as an acid—base neutralization reaction. v“ Recisonobie Answer Check: The reactions that ar'e not oxidation—reduction reactions ar'e easily classified as one of the other reactions we have studied in this chapter. 79. Ann-er: 121mL HNO3 Sn‘nregr and Explanation: Given the mass ofone reactant. the balanced chemical equation for a reaction. and the molaiity of a solution containing the other reactant. deteimine the volume of the second solution for a complete reaction. The stoichiometi}! of a balanced chemical equation dictates how the moles of reactants combine. so we will connnonly look for ways to calculate moles. Here. the mass and molar mass can be used to find the moles of one reactant. Then we will use the equation stoichionieti}! to find out moles of the other reactant needed. Then we will use the moles and morality to find volume in liters and convert liters into milliliters. Notice: It is NOT appropriate In use the (if ..:'on equation Wile??? uprising trio"? reactions." "We learn from the balanced equation that 1 mol of Battll-I)2 reacts with 2 mol 1-1303. ' ‘ . 7 ' t . 1.30 g Baiomj X 1 mol BatOl—lih. X -mol I-_‘:\.O_. ‘ ' 1?].3416 g Baron)1 111101 BatOH); X l L l—TND; solution x 1000 mL [HES moi lib—O3 1L — 121 mL HNO; solution .99. Answer: The only redox reaction is reaction (c); oxidizing agent is Ti; reducing agent is Mg. Sh'ategr and Exp-Taxation: Given several reactions. determine if they are redox (oxidation—reductionj reactions and identify the oxidizing and reducing agents in those that are redox reactions. To decide if a reaction is a redox reaction. we need to see if any of the elements change oxidation state. In redox reactions. the atoms change oxidation states during the reaction. If no change in oxidation state is observed. then the reaction is not a redox reaction. The oxidizing agent is the reactant that assists an oxidation by being reduced. so it will be the reactant whose atoms gain electrons. and end up with a lower (more negative or less positive} oxidation number. The reducing agent is the reactant that assists a reduction by being oxidized so it will. be the reactant whose atoms lose electrons. and end up with a higher (more positive or less negative} oxidation number. (a) Look at the oxidation numbers for the reactants: NaOH contains a monatornic cation. Na‘. and a diatomic anion. OH‘. Rule 2 gives us Ox. s" Na = —1. Rule 3 gives us Ox. e‘ O = —2 and 0x. 5? 1-1 = —1. H3PO_. contains oxygen and hydrogen. Rule 3 gives us Ox. if 0 = —2 and Ox. # H = -1. We use the sum rule to find the Ox. #5' P. For a molecule. the sun] of the oxidation numbers is equal to zero. O=3><(+l)-1xfOx.#P)—4xI:—2} Therefore. Ox. # P = —5. Look at the oxidation numbers for the products: NaHJPO4 contains a nionatomic cation. Na'. and a polyatomic anion. HEPOf. Rule 2 gives us Ox. :3' Na = —l. HEPOJ‘ contains oxygen and hydrogen. Rule 3 gives us Ox. if 0 = —2 and Ox. if H = +1. We use the sum rule to find the Ox. e‘ P. For a polyatomic anion. the sum ofthe oxidation numbers is equal to the ions charge of l—. —1 =2 xf—l)—1x(Ox..='=P:I—4 x (—2} Therefore. Ox. # P = —5. H10 contains oxygen and hydrogen. Rule 3 gives us Ox. e‘ O = —3 and Ox. :3' H = —l. The oxidation numbers don't change. so this is NOT a redox reaction. (b) Look at the oxidation numbers for the reactants: PCT-l3 contains hydrogen. Rule 3 gives us Ox. .i.‘ H = —1. We use the sum rule to find the Ox. e‘ N. For a molecule. the sum of the oxidation numbers is equal to zero. 0 =1 xfOX.#N)—3 xf-lj Therefore. Ox. e N = —3 CO3 contains oxygen. Rule 3. gives us Ox. :3' O = —2. We use the sum rule to find the Ox. :3' C. For a molecule. the sum of the oxidation numbers is equal to zero. 0:1 mmeq—z xf—l) Therefore. Ox. # C‘ = —4. H10 contains oxygen and hydrogen. Rule 3 gives us Ox. e‘ O = —2 and Ox. :3' H = —l. Look at the oxidation numbers for the products: Nl-IfiJ-IC‘O3 contains a polyatomic cation. Ell-14‘. and a polyatonuc anion. HCOS‘. INTI-If contains hydrogen. Rule 3 gives us Ox. if H = —1. We use the sum rule to find the Ox. 9': N. For a polyatomic cation. the sum of the oxidation numbers is equal to the ions charge of 1+. —l = l >< (Ox. #N't—slx'fil) Therefore. Ox. # N = —3. HC'OE‘ contains oxygen and hydrogen. Rule 3 gives us Ox. #f O = —2 and Ox. e” H = —l. “We use the sum rule to find the Ox. :3' C. For a polyatonuc anion. the sum of the oxidation numbers is equal to the ions charge of l—. —1=l><(-l:I—l><(Ox..'=.' C) —3 x (—3] Therefore. Ox. e C = —4. The oxidation numbers don't change. so this is NOT a redox reaction. (c) Look at the oxidation numbers for the reactants: TiCl4 contains a nionatomic cation. Ti“. and a nionatornic anion. Cl‘. Rule 2 gives us Ox. e Ti = —4 and Ox. .t—r' C1=—l. Rule 1 indicates the oxidation number of elements is zero. so Ox. # Mg = 0. Look at the oxidation numbers for the products: Rule 1 indicates the oxidation number of elements is zero. so Ox. 7—? Ti = 0. MgClg contains a monatornic cation. Mg}. and a monatomic anion. (‘1‘. Rule 2 gives us Ox. :3' Mg = —2 and Ox. .-"—." Cl =—l. The oxidation numbers of Mg and Ti do change. so this IS a redox reaction. The oxidizing agent is Ti since its Ox. Ii goes from +4 to zero. ' n. The reducing agent is Mg since its Ox. if goes front zero to —l. (d) Look at the oxidation numbers for the reactants: NaC‘l contains a nionatomic cation. Na'. and a monatomic anion. C‘l‘. Rule 2 gives us Ox. e‘ Na = —l and Ox. .'=.' C1=—l. NaI-ISOJ contains a monatomic cation. Na". and a polyatomic anion. H804". Rule 2 gives us Ox. .'=.' Na = —1. H504‘ contains oxygen and hydrogen. Rule 3 gives us Ox. e O = —2 and Ox. 5? H = —1. Wis-"e use the sum rule to find the Ox. e S. For a polyatomic anion. the sum of the oxidation numbers is equal to the ions charge of 1—. —1 =1 x (—1ji— l X (om! .s)—4 x (—2) Therefore. Ox. # S = —6. Look at the oxidation numbers for the products: l—IC'l contains hydrogen. Rule 3 gives us Ox. .'=.' H = —1. We use the sum rule to find the Ox. e‘ ('1. For a molecule. the sum of the oxidation numbers is equal to zero. O=l><(+1:l-1><(OX.#C1) Therefore. Ox. # C'l = —l. NasSO4 contains a nionatontic cation. Na". and a polyatomic anion. 3043‘. Rule 2 gives us Ox. #5' Na = —l. 5043‘ contains oxygen. Rule 3 gives us Ox. 5'4 0 = —2. We use the sum rule to find the Ox. is S. For a polyatomic anion. the sum of the oxidation numbers is equal to the ions charge of 2—. —2= 1 ><(Ox.#Sft—4><(—2) Therefore. Ox. e S = —6. The oxidation numbers don't change. so this is NOT a redox reaction. v“ Reosorsobfe Answer Check: The oxidation numbers are consistent with typical oxidation states of these elements. The reactants and products are common acids. ionic compounds and ions. ...
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This note was uploaded on 06/03/2008 for the course CHEM 1A taught by Professor Nitsche during the Fall '08 term at University of California, Berkeley.

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MOOREHW4 - Chapter 4: Quantities of Reactauts and Products...

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