Chemistry 1A3 Instructor: R.S. Dumont Supplementary Problem Solutions Types of Reactions

Chemistry 1A3 Instructor: R.S. Dumont Supplementary Problem Solutions Types of Reactions 1. (a) 2 Ca(s) + O 2 (g) 2CaO(s) (b) 2 Na(s) + Cl 2 (g) 2NaCl(s) (c) 16 Ga(s) + 3 S 8 (s) 8Ga 2 S 3 (s) (d) 3 Ba(s) + N 2 (g) Ba 3 N 2 (s) (e) C(s) + O 2 (g) CO 2 (g) (f) 12 K(s) + P 4 (s) 4K 3 P(s) 2. Everything has some solubility, though it can be very small. Thus, some glass (a very tiny concentration) will dissolve in water. This solubility is increased to its largest value at the boiling point of the water. The dissolved glass will redeposit itself on the inner surface of the flask. When it does so, it appears as a non-smooth residue. Several days of boiling will produce a sufficient quantity of residue to be visible to the early alchemists. Thus, all they really accomplished was a transformation of the inner surface of the glass flask. 3. (a) KCl(s) H 2O K + (aq) + Cl (aq) (b) MgI 2 (s) H 2O Mg 2+ (aq) + 2 I (aq) (c) HBr(g) H 2O H + (aq) + Br (aq), or HBr(g) + H 2 O(l) H 3 O + (aq) + Br (aq) resulting in an acidic solution. (d) FeSO 4 (s) H 2O Fe 2+ (aq) + SO 2 4 (aq) (e) Ca(OH) 2 (s)isinsoluble. However,totheextentthatitdoesdissolve, the breakup occurs as Ca(OH) 2 (s) H 2O Ca 2+ (aq) + 2 OH (aq) resulting in a slightly basic solution. 1

(f) NaCH 3 COO(s) H 2O Na + (aq) + CH 3 COO (aq). A subsequent hydrolysis reaction, CH 3 COO (aq) + H 2 O(l) *) CH 3 COOH(aq) + OH (aq) produces a basic solution. (g) H 2 SO 4 (l) H 2O H + (aq) + HSO 4 (aq) or H 2 SO 4 (l) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) resulting in an acidic solution. HSO 4 is itself a weak acid. Thus, a subsequent dissociation occurs (to a small extent); HSO 4 (aq) *) H + (aq) + SO 2 4 (aq) (h) LiOH(s) H 2O Li + (aq) + OH (aq) resulting in a basic solution. 4. (a) (NH 4 ) 2 SO 4 (aq) + Ba(NO 3 ) 2 (aq) BaSO 4 (s) ppt +2NH + 4 (aq) + 2 NO 3 (aq) ions remaining in solution (b) 2 NaCl(aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s) +2Na + (aq) + 2 NO 3 (aq) ppt (c) Na 3 PO 4 (aq) + KNO 3 (aq) no ppt (Na +,K +,PO 3 4 and NO 3 all remain in solution) (d) LiBr(aq) + RbCl(aq) no ppt (Li +,Rb +,Br and Cl all remain in solution) (e) CuCl 2 (aq) + 2 NaOH(aq) Cu(OH) 2 (s) +2Na + (aq) + 2 Cl (aq) ppt (f) Hg 2 (NO 3 ) 2 (aq) + CaCl 2 (aq) Hg 2 Cl 2 (s) +Ca 2+ (aq) + 2 NO 3 (aq) ppt 5. Since no precipitate formed when an aqueous solution of NaCl or Na 2 SO 4 was added, the solution cannot contain Hg 2+ 2, which precipitates with both Cl and SO 2 4 (see note below), or Ba 2+, which precipitates with SO 2 4. The formation of a precipitate with OH is consistent with the presence of Mn 2+.Thus,onlyMn 2+ is present in the sample solution. Note - Table 4.2 in Chang does not list the insolubility of Hg 2 SO 4. You are therefore not required to know this fact. Hg 2 SO 4 is nevertheless insoluble. However, you do not need this knowledge to answer the question. 2

6. (a) 2 HClO 4 (aq) + Mg(OH) 2 (s) Mg(ClO 4 ) 2 (aq) + 2 H 2 O(l) 2H + (aq) + Mg(OH) 2 (s) Mg 2+ (aq) + 2 H 2 O(l) (b) CH 3 COOH(aq) + NaOH(aq) NaCH 3 COO(aq) + H 2 O(l) CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) (c) H 2 SO 4 (aq) + 2 NH 3 (g) (NH 4 ) 2 SO 4 (aq) 2H + (aq) + 2 NH 3 (g) 2NH + 4 (aq) 7. 50.00 ml = 0.05000 L of 0.300 M NaOH = (0.05000 L) ³ 0.300 mol L 1 = 0.0150 mol of OH (a) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) or H + (aq) + OH (aq) H 2 O(l) (b) (c) = we need 0.0150 mol of H + to react completely with 0.0150 mol of OH V = V = = ³ 0.100 mol L 1 V = 0.0150 mol or 0.0150 mol = 0.150 L = 150 ml of HCl solution 0.100 mol L 1 0.0150 mol 0.150 mol L = 0.100 L = 100 ml of HNO 1 3 solution 0.0150 mol V = 0.200 mol L = 0.075 L = 75 ml of CH 3COOH solution 1 Note that CH 3 COOH has only one acidic H - the one bonded to an O. See question 1(b). 8. (a) 2NaN 3 (s) 2Na(s)+3N 2 (g) (1) 3

(b) 5 ³ 2Na Na 2 O+2e 1 ³ 2KNO 3 +10e K 2 O+N 2 10 Na(s) + 2 KNO 3 (s) K 2 O(s) + 5 Na 2 O(s) + N 2 (g) Note that everything but the O s is balanced in the half reactions. You cannot use the rules for balancing O s (& H s) for aqueous acid or base solution, since this is a solid state reaction - with a gaseous product. Miraculously, the O s balance in the final equation. To determine the required amount of sodium azide, we must first determine the desired number of moles of N 2 (g). Assume N 2 is an ideal gas at 22 C and 1 atm pressure (a good assumption). In this case, n = pv RT = (1 atm) (70 L) (0.082 L atm K 1 mol 1 ) (295 K) =2.89 4 mol Reaction 1 shows that x mol of NaN 3 produces x molofnaand 3 2 x mol of N 2.Reaction?? shows that the x molofnagoesonto produce an additional 1 10 x mol of N 2. This last statement assumes there is sufficient KNO 3 to consume all of the sodium - potassium nitrate is present in excess in car airbags. Overall, x mol of NaN 3 leads to 1.6x mol of N 2. To get 2.89 4 mol of N 2 we need 9. (a) CH 3 COOH(aq) (A) CH 3 COOH(aq) (A) 1.6x =2.89 4 or x =1.80 8 = (1.80 8 mol) ³ 65.02 g mol 1 =118gofNaN 3. +NaNH 2 (s) (B) + NaNH 2 (s) (B) NaCH 3 COO(aq) + NH 3 (aq) (AB) Na + (aq) + CH 3 COO (aq) + NH 3 (aq) Note that a subsequent reaction can take place between unreacted CH 3 COOH(aq) and the product, NH 3 (aq). The following net ionic equation describes the overall reaction (consisting of two steps): 2CH 3 COOH(aq) + NaNH 2 (s) Na + (aq) + NH + 4 (aq) + 2 CH 3 COO (aq) 4

(b) F 2 (g) + 2 NaBr(aq) 2NaF(aq)+Br 2 (aq) (OR) F 2 (g) +2Br (aq) 2F (aq) + Br 2 (aq) (c) Sr(s) +2HI(aq) SrI 2 (aq) + H 2 (g) (OR) Sr(s) +2H + (aq) Sr 2+ (aq) + H 2 (g) (d) AgClO 4 (aq) + LiCl(aq) AgCl(s) ppt + LiClO 4 (aq) (P) Ag + (aq) + Cl (aq) AgCl(s) ppt (e) Al(OH) 3 (s) (A) +3HNO 3 (aq) (B) Al(NO 3 ) 3 (aq) + 3 H 2 O(l) (AB) Al(OH) 3 (s) (A) +3 H + (aq) (B) Al 3+ (aq) + 3 H 2 O(l) (f) NaCl(aq) + Br 2 (aq) NR (g) 2 K(s) +2H 2 O(aq) 2KOH(aq)+H 2 (g) (OR) (h) Mg(s) 2K(s) equation +Cl 2 (g) +2H 2 O(aq) 2K + (aq) + OH (aq) + H 2 (g) MgCl 2 (s) (OR) also the net ionic 10. Species present in solution are denoted by species (a) NH 3 (g) H 2O A basic solution. NH 3 (aq) aweakbase= hydrolysis NH 3 (aq) + H 2 O(l) *) NH+ 4 (aq) + OH (aq) 5

(b) NH 4 I(s) H 2O NH + 4 (aq) + I (aq) acid conjugate to a weak base = hydrolysis NH + 4 (aq) + H 2 O(l) *) NH 3(aq) + H 3 O + (aq) An acidic solution. (c) KI(s) H 2O K+ (aq) + I (aq) (d) ClO 4 H(l) H 2O H+ (aq) + ClO 4 (aq) Of course, H + (aq) is really just shorthand for H 3 O + (aq). The process could be written ClO 4 H(l) + H 2O(l) H 3 O + (aq) + ClO 4 (aq) An acidic solution. Note there is (essentially) no ClO 4 H(aq) because ClO 4 H is a strong acid - perchloric acid. (e) NaCH 3 COO(s) H 2O Na + (aq) + CH 3 COO (aq) base conjugate to weak acid = hydrolysis CH 3 COO (aq) + H 2 O(l) *) CH 3COOH(aq) + OH (aq) A basic solution. 11. I have labeled the oxidizing and reducing agents on the right in order to emphasize that the products of a redox reaction are just another pair of oxidizing and reducing agents. (a) ON=0 Ni (s) +2 ON=+1 H Cl(aq) ON=+2 Ni 2+ (aq) + ON=0 H 2 (g) +2Cl (aq) (b) SiCl 4 (l) + 2 H 2 O(l) 4HCl(aq)+SiO 2 (s) NOT aredox reaction (c) ON=0 O 3 (g) + ON=+2 N O(g) O 2 (g) + ON=+4 N ON= 2 O 2 (g) This reaction is already balanced. (d) 3 ³ Cu Cu 2+ +2e 6

2 Ã! ON=+5 N O 3 +3e +4H + ON=+2 N O + 2 H 2 O 3 Cu(s) +2 NO 3 (aq) +8H + (aq) 3 Cu 2+ (aq) +2 NO(g)+4H 2 O(l) (e) ON=+6 Cr 2 O 2 7 +14H + +6e 2 ON=+3 Cr 3+ +7H 2 O Ã 3 2 ON= 1! Cl ON=0 Cl 2 +2e Cr 2 O 2 7 (aq) +6 Cl (aq) +14H + (aq) 2 Cr 3+ (aq) +3 Cl 2 (g) +7H 2 O(l) (f) 2 ON=+3 N O 2 +5H 2 O+6e ON= 3 N H 3 +7OH Ã ON=0 Al + 4 OH ON=+3 Al O 2 +2H 2 O+3e! NO 2 (aq) +2 Al(s) +H 2 O(l) + OH (aq) NH 3 (g) +2 AlO 2 (aq) 12. 5.00 g of calcium metal = 5.00 g 40.08 g mol 1 = 0.124 8 mol of Ca which produces 0.249 6 mol of OH (aq) via Ca(s) + 2 H 2 O Ca 2+ (aq) + 2 OH (aq) + H 2 (g). Making the final volume up to 500 ml = 0.500 L yields a solution with [OH ]= 0.249 6 mol =0.499 M. 0.500 L 13. 17.25 ml = 0.01725 L of 0.216 M KMnO 4 = (0.01725 L) ³ 0.216 mol L 1 = 0.00372 6 mol of MnO 4 7

which react with 5 2 0.00372 6 =0.00931 5 mol of H 2 C 2 O 4 via 5H 2 C 2 O 4 (aq) + 2 MnO 4 (aq) + 6 H+ (aq) 2Mn 2+ (aq) + 10 CO 2 (g) + 8 H 2 O(l) 0.00931 5 mol of H 2 C 2 O 4 resulted from 0.00931 5 mol of C 2 O 2 4 which came from 1.356 g of the new iron compound. In case of K 3 [Fe(C 2 O 4 ) 3 ], there are 3 mol of C 2 O 2 4 per mol of compound. If the new compound were K 3 [Fe(C 2 O 4 ) 3 ], then the formula weight would be 1.356 g 1 0.00931 3 5 mol = 436. 7 gmol 1. In case of K[Fe(C 2 O 4 ) 2 (H 2 O) 2 ], there are 2 mol of C 2 O 2 4 per mol of compound. If the new compound were K[Fe(C 2 O 4 ) 2 (H 2 O) 2 ], then the formula weight would be 1.356 g 1 2 0.00931 5 mol = 291. 1 gmol 1. Clearly the unknown copound is K 3 [Fe(C 2 O 4 ) 3 ], since only in this case is the calculated formula weight consistent with the molecular formula. Formula weights: Compound K 3 [Fe(C 2 O 4 ) 3 ] K[Fe(C 2 O 4 ) 2 (H 2 O) 2 ] Formula Weight 437.2 307.0 14. (a) Ca 2+ (aq) + CO 2 3 (aq) CaCO 3 (s) (P) (b) 2 Li(s) (c) F 2 (g) +2H 2 O(l) +2Br (aq) 2Li + (aq) + H 2 (g) + 2 OH (aq) 2F (aq) + Br 2 (aq) (d) Sr(s) +2H + (aq) Sr 2+ (aq) + H 2 (g) (OR) There is a subsequent precipitation reaction: (OR) (OR) (e) HCO 3 (aq) & (A) &(AB) Sr 2+ (aq) + SO 2 4 SrSO 4 (s) (P) +NaH(s) CO 2 3 (aq) + Na + (aq) + H 2 (g) (OR) & (B) 8

(f) LiHCO 3 (s) +H + (aq) Li + (aq) + H 2 CO 3 (aq) (AB) (B) (A) There is a subsequent (gas-forming) reaction of H 2 CO 3 (aq): H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) (g) Hg 2 (NO 3 ) 2 (aq) + NaClO 3 (aq) NR (h) H + (aq) + OCl (aq) HOCl(aq) 15. (a) Sodium hydrogen sulfite plus vinegar: NaHSO 3 (s) + CH 3 COOH(aq) NaCH 3 COO(aq) + H 2 SO 3 (aq) H 2 SO 3 (aq) H 2 O(l) + SO 2 (g) (gas-forming) In the case of sodium hydrogen carbonate plus vinegar, NaHCO 3 (s) + CH 3 COOH(aq) NaCH 3 COO(aq) + H 2 CO 3 (aq) H 2 SO 3 (aq) H 2 O(l) + SO 2 (g) (gas-forming) (acid-base) (acid-base) Net ionic equations are ALSO accepted. However, because NaHSO 3 and NaHCO 3 aresolids,theonlydifference in the net ionic equations is the separate listing of Na + (aq) & CH 3 COO (aq) instead of NaCH 3 COO(aq). (b) 10.0 g of SO 2 (g) corresponds to 10.0 g 64.07 g mol 1 =0.156 1 mol (2) (Note, the notation, 0.156 1 givestheextra-notsignificant digit - as a subscript. The idea is to carry an extra digit through subsequent calculations to reduce roundoff error. Only significant digits are given in the final answer.) To produce 10.0 g of SO 2 (g), we need 0.156 1 mol of H 2 SO 3, which in turn requires 0.156 1 mol of NaHSO 3,or (0.156 1 mol) ³ 104.07 g mol 1 = 16.2 g. (3) 9

(c) 0.156 1 mol of CH 3 COOH is consumed. = (0.156 1 mol) ³ 60.05 g mol 1 =9.37g 16. (a) (d) 10.0 g of NaHCO 3 = 10.0 g of CH 3 COOH = 10.0 g 84.01 g mol 1 =0.119 0 mol 10.0 g 60.05 g mol 1 =0.166 5 mol NaHCO 3 is clearly the limiting reagent (fewer moles in a reaction with equal stoichiometries). = 0.119 0 mol of CO 2 (g) is produced, all of the NaHCO 3 is consumed, and 0.166 5 0.119 0 =0.048molofCH 3 COOH remains unreacted. The remaining solution is clearly acidic. 3 I (aq) 3I I 3 +2e Cl 2 +2e + Cl 2 (aq) 2Cl I 3 (aq) + 2 Cl (aq) (b) Ca(OH) 2 (s) + 2 HNO 3 (aq) Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) An acid-base reaction. (c) This one is tricky. It is most easily balanced algebraically as follows: Let a KO 2 (s) + b H 2 O(g) c KOH(s) + d O 2 (g) 10

be the balanced reaction, where a, b, c and d aretobedetermined. Because H appears in only one compound on each side of the reaction, we must have 2b = c. Similarly, balancing K atoms gives Thus, a = c. 2b KO 2 (s) + b H 2 O(g) 2b KOH(s) + d O 2 (g). Now we can balance O atoms to get 4b + b =2b +2d or 2d =3b. The smallest whole numbers which solve this equation are which gives b =2 and d =3 4 KO 2 (s) +2 H 2 O(g) 4KOH(s)+3O 2 (g) (d) CH 3 C H 2OH CH 3 C OOH + 4 ON= 1 ON=+3 e (4) Mn ON=+7 O 4 (aq) + 5 e Balance (4) in acid: Mn 2+ (aq) (5) ON=+2 CH 3 CH 2 OH + H 2 O CH 3 COOH + 4 H + +4e (6) Balance (5) in acid: MnO 4 +8H + +5e Mn 2+ +4H 2 O (7) Balance e s: 5 (6) + 4 (7) gives 5 CH 3 CH 2 OH(aq) +4 MnO 4 (aq) +12H + (aq) 5CH 3 COOH(aq) + 4 Mn 2+ (aq) + 11

(e) H 2 2H + +2e 2 ³ Fe 3+ +e Fe 2+ (f) H 2 (g) +2 Fe 3+ (aq) 2H + (aq) + 2 Fe 2+ (aq) µ 2 Mn ON=+7 O 4 +2H 2O+3e Mn O 2 +4OH ON=+4 2 MnO 4 (aq) µ 3 Zn ON=0 +2OH Zn (OH) 2 +2e ON=+2 +3 Zn(s) +4H 2 O(l) 2MnO 2 (s) + 3 Zn(OH) 2 (s) + 2 OH (aq) 17. (a) NH 4 NO 3 (s) is an ionic solid consisting of NH 4 ions and NO 3 ions. N ON= 3 H 4 N O 3(s) heat N 2 ON=+5 ON=+1 O(g) + 2 H 2 O(g) (b) 1 kg = 1000 g of NH 4 NO 3 = 1000.0 g = 12.5 mol 80.044 g mol 1 Note that, since I gave only one significant digit in the mass of ammonium nitrate (1 kg), only the 1 in 12.5 mol is significant. Each mole of NH 4 NO 3 produces 1 mol of N 2 O(g) and 2 mol H 2 O(g); i.e. 3 mol of ideal gas in total. Therefore, we have 3 12.5 = 37.5 mol of ideal gas. According to the ideal gas law, V = nrt /p = (37.5 mol) ³ 0.082 L atm K 1 mol 1 (298 K) / (1 atm) = 916 L = 900 L. In the last step, I round the result to 1 significant digit. 12

18. (a) 2 Co 3+ (aq) +2Cl (aq) 2Co 2+ (aq) +Cl 2 (aq) (b) PbSO 4 (s) (c) Br 2 (aq) = Co 3+ (aq) is a stronger oxidizing agent than Cl 2 (aq) +Cd(s) Pb(s) +Cd 2+ (aq) = PbSO 4 (s) is a stronger oxidizing agent than Cd 2+ (aq) +Pb(s) +SO 2 4 (s) 2Br (aq) + PbSO 4 (s) = Br 2 (aq) is a stronger oxidizing agent than PbSO 4 (s) (d) Cl 2 (g) +2Br (aq) 2Cl (aq) +Br 2 (aq) = Cl 2 (g) is a stronger oxidizing agent than Br 2 (aq) We assume that Cl 2 (g) and Cl 2 (aq) have essentially the same oxidizing strength. This permits ordering of all the oxidizing agents according to oxidizing strength: Co 3+ (aq) > Cl 2 (aq) > Br 2 (aq) > PbSO 4 (s) > Cd 2+ (aq) Similar reasoning leads to the following ordering of reducing agents according to reducing strength: Co 2+ (aq) < Cl (aq) < Br (aq) < Pb(s) < Cd(s) 13