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Chem2090-PS3Key - PflafiLg/M 3E7 3(5/7 4.6 4.1(a...

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Unformatted text preview: PflafiLg/M 3E7 3 [(5/7 4.6. 4.1; (a) m0102=32.8gKC103x j ”:60. (1)) mass KClO3 = 50.0 g 02 x (c) mass KCl = 28.3g 02 x ‘ £1.64. ..,-—-— J (a) 3 Mg (5) + N2(g)—)Mg31<12(s) (b) K0103 (s)—>KCl(s) + zoz (g) or 2 KC103(s)—->2 Kc1(s) + 3 02 (g) (c) NaOH(s) + NH4Cl(s)————>NaCl(s) + NH3 (g) + H20(g) (d) 2 Na(s) + 2H2 0(1) —> 2 NaOH(aq) + H2 (g) 1 mol KClO 3 mol 02 122.6 g KClO3 x2 mol KClO3 lmol 02 x 2molKClO3 x122.6g KCIO,_ =128gKC103 32.00g Oz 3mol O2 lmol KClO3 1m9102 x 2mol KCl x 74.55g KCl = 44.0g KCl 32.00g O2 3mol O2 lmol KCl — —O.401mol 02 Since the two reactants combine in an equimolar basis, the one present with the fewer number of moles is the limiting reactant and determines the mass of the products. 1L 0.275 mol ZnSO mol ZnSO4 = 315 mL x —————— x -—————_,——i = 0.0866 mOl ZnSO4 1000 mL 1L soln mol BaS = 285mL>< 11‘ 0315mm 335 = 0.0898mol BaS 1000mLx lLsoln . . ‘~ Thus, ZnSO4 is the limiting reactant and 0.0866 mol of each of the products will be produced. mass products = 0.0866 mol Baso. x W+ +0 0866 mol ZnS x _'___97 463 Zns lmol B3804 k lmol Zns = 28.7 g product mixture (lithopone) lmol C 6HHOH lmol C‘SH10 100.16g C 6HHQHx lmol CfiHUOH (a) mass 06H“, =100.0g CéHllOHx x 82.146g CfiHl . lmol CGHm' _ 64. 0 gC6 H10 produced 82. 01 g C 6H10 calculated (c) mass C HHOH— = 100 0 g c Hm produced x Wx M 0. 780g produced 82. 15g C6 Hl0 xlmolC GHHOHX 100 2gC HHOH lmolC 610H lmolC 6HHOH ° = 82.01 g C6H10 = theoretical yield (b) percent yield— x.100%=7s 0% yield = 156g CéHnOH are needed NaI(aq)+ AgN03(aq) —> AgI(s )+ NaN03(aq) (multiply by 4) 2 AgI(s) + F e(s) —) F eI2(aq) + 2 Ag(s) (multiply by 2) 2 Feb a + 3 C12 —~> 2 FeCl; a + 2 I2 3 unchan ed 4Nal(aq) + 4AgN03(aq) + 2Fe(s) + 3CI2(g)—> 4NaNO3(aq) + 4Ag(s) + 2FeC13(aq) + 212(3) For every 4 moles of AgNO3, 2 moles of I2(s) are produced. The mass of AgNO; required = 1.00 kgI2(s)x1000 g I 2x(s) 1 molI 2(s) x 4 mol AgNO3 (s) x 169.873 g AgNO3 (s) 1 kg12(s) x.253 809 g12(s) 2 molI 2(5) 1 mol AgNO3 (s) = 1338. _5_9 g AgNO3 per kg of 12 produced or 1.34 x 103 g AgN03 per kg of 12 produced 5‘, L. 9 (a) Because its formula begins with hydrogen, HCSHSO is an acid. it is not listed in Table 5—1, so it is a weak acid. A weak acid is a weak electrolyte. (b) LiZSO4 is an ionic compound, that is, a salt. A salt is a strong electrolyte. (c) MgI2 also is a salt, a strong electrolyte. (d) (CHaCH2 )2 O is a covalent compound whose formula does not begin with H. Thus, it is neither an acid nor asalt. It also is not built around nitrogen, and thus it does not behave as a weak base. This is a nonelectrolyte. (e) Sr (OH)2 is a strong electrolyte, one of the strong bases listed in Table 5-1. 5'. 2_1_. Mixture Result (net ionic equation) (a) HI(a) + Zn(NO3 )2 (aq): No reaction occurs. (b) CuSO4 (aq) + Na2C03 (aq): Cu2+ (aq) + CO:' (aq) —> CuCO3 (s) (c) Cu (N03 )2 (aq) + Na3P04 (aq): 3Cu2+ (aq)+ 2PO43" (aq) —> Cu3 (Po, )2 (s) S , 38. (a) Oxidation: {P4 (s)+16 H200) ——) 4 HZPO; (aq) + 24 H+ (aq) + 20 e‘ } x 3_ Reduction: {NOa' (aq) + 4 H+ (aq) +3 6‘ —> N0(g)+ 2 H200) } x 20 Net: 3 P,(s)+20 No;(aq)+8 H20+8 H*(aq)~>12 H,i>0,‘(aq)+2o N0(g) (b) Oxidation: {szo,z‘(oq)+‘5 H,O(1);>2 so,” (aq)+10 H*(aq)+3 e' }’x‘5 Reduction: {Mn04_(aq)+8 H*(aq)+5 e' —) Mn2*(aq)+4 H20(l) }x 8 Net: 5 s20," (aq)“kg Mno; (aq)+14 H*(aq) —+ 10 so," (iq)+3 Mn2+ (aq)+7 H200) (c) Oxidation: 2 HS‘ (aq)+3 HZO(1')-> 82032“ (aq)+8 H” ‘(aq)+,8 e' Reduction: {2 Hso,‘(aq)+4 ‘H+(aq)+4 e‘ —+ s,o,z*(aq)+3 H200) }x 2 Net: 2 HS‘(aq)+4 Hso,‘(aq)—)3 SZOJZ‘(aq)+3 H200) (d) Oxidation: 2 NH3OH“ (aq) —> N,0(g) + H200) + 6 H+ (aq) fr 4 6" Reduction:'{ Fe3+ (aq) + e‘ —-> Fe2+ (3C1) . i } x 4 Net: 4 Fe“ (aq) + 2 NI-II3OH+ (aq) ——) 4 Fe2+ (aq) + N20 (g) + H20(1)+6 H+ (3C1) 5‘; 51.. (a) Oxidation: C12(g)+12 OH’ (aq)—)2 C10; (aq)+6 H20(l)+10 e~ Reduction: {C12 (g)+2 e’ —> 2 Cl" (aq) } x 5 Net: 6 c12(g)+12 OH'(aq)—)10 Cl'(aq)+2 C103'(aq)+6‘*H20(1) Or: 3 C12(g)+6 0H”(aq)—>5 Cl‘(aq)+ Clo;(aq)+3 H200) (b) Oxidation: SZO4Z_(aq)+2 H20(l)—>2 Hso;(aq)+2 H+(aq)+2 6' Reduction: SZO4Z_(aq)+2 H+(aq)+2 e- —+ szo3z‘(aq)+ H200) Net: 2 SZO4z'(aq)+ H20(l)—>2 Hso;(aq)+ 32032_(aq) 5:351. First balance the titration equation: Oxidation: {C2042" (aq) ~—> 2 CO2 (g) + 2 e" } x 5 Reduction: {MnOf (aq)+8 .H* (aq)+5 e" 4+ Mn“ (aq)+4 H200) }x 2 Net: 5 0203' (aq)+2 Mno; (aq)+16 H+ (aq) —+ 10 co2 (g)+2 Mn2+ (aq)+8 H20(1) 1000 mL x 25.8 mL satd soln KMnO4, x 0.02140 mol KMnO4 I L 5.00 mL satd soln NazCZO4 1000 mL KMnO4 x l moanO4' x 5 molCZOf‘ x l molNa2C204 x 134.0 gNa2C204 1 molKMnO4 2 moanO; V l molCZOf' 1 molNaZC204 massmzczo‘ = 37.0 gNaZCzQ‘ massmzczo‘ =1.00 L satd soln NaZC204 x ...
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