SOLUTIONS TO PRACTICE PROBLEMS
SOLUTIONS TO PRACTICE PROBLEMS
1. (1.00tonFe)(2000lb/ton)(453.6g/lb)(moleFe/55.85gFe)((moleFe3O4)/3moleFe)(232g Fe3O4/mole Fe3O4)(100.0g ore/2.50g Fe3O4)(lb/453.6g)(ton/2000lb) = 55.3 ton ore
2. 4FeS2 + 11O2(g) → 8SO2(g) + 2Fe2O3 2SO2(g) + O2(g) → 2SO3(g)
SO3(g) + H2O → H2SO4
a. (1.00ton soln)(.98ton H2SO4/ton soln)(2000lb/ton)(453.6g/lb)(mole H2SO4/98.1g) (mole SO3/mole H2SO4)(mole SO2/mole SO3)(4mole FeS2/8mole SO2) =
= 4.53 x 103 mole FeS2
(4.53 x 103 mole FeS2)(120.0gFeS2/mole)(100.0g ore/15.0gFeS2)(lb/453.6g) (ton/2000lb) = 4.00 ton ore
b. O2 is required in two steps
(4.53x103 mole FeS2)[(11mole O2/4 mole FeS2)+(8mole SO2/4moleFeS2)(moleO2/
2mole SO2)](100.0mole air/21.0mole O2)(.08205Latm/moleK) (298.15K/620.0mmHg)(atm/760.0mmHg) = 2.43x106 L air.
3. 3Cu2S 16H3O+ +10 NO3- → 6Cu2+ + 3SO42- + 10NO(g) + 24 H2O
3Cu2+ + 2 Fe → 3Cu + 2Fe3+
a. (1.00ton ore)(12.0tonCu2S/100.0ton ore)(2000lb/ton)(453.6g/lb)(moleCu2S/ 159.15g)(6mole Cu2+/3mole Cu2S)(2moleFe/3moleCu2+)(55.85gFe/mole)
(lb/453.6g)(ton Fe/2000lb) = 0.0561 ton Fe
b. (1.00ton ore)(12.0tonCu2S/100.0ton ore)(2000lb/ton)(453.6g/lb)(moleCu2S/ 159.15g)(10moleNO/3moleCu2S)(.08205Latm/moleK)(300.15K/620.0mmHg)
(760.0mmHg/atm) = 6.88x104 L NO(g)
4. 2Ni3S2 + 7O2(g) → 6NiO + 4SO2(g) NiO + C → Ni + CO(g)
a. (kg Ni3S2)(1000g/kg)(mole Ni3S2/240.2g)(3moleNi/mole Ni3S2)(58.7gNi/mole) =
736gNi
c. (kg Ni3S2)(1000g/kg)(mole Ni3S2/240.2g)(4moleSO2/2moleNi3S2)(.08205Latm/
moleK)(294.15K/620mmHg)(760mmHg/atm) = 246 L SO2
5. Since Ni → Ni2+ in Ni(NO3)2 and Cu → Cu2+ each has 2 equivalents per mole
(100g alloy)/[(67.0gNi)(mole/58.7gNi)(2eq/moleNi) + (28.0gCu)(mole/ 63.5gCu) (2eq/moleCu)] = 31.6 g alloy/eq
6. Ni 4CO(g) → Ni(CO)4(g)
(5.60kgCO)(1000g/kg)(moleCO/28.0gCO)(moleNi(CO)4/4moleCO) = 50.0 moleNi(CO)4
(1470gNi)(moleNi/58.7gNi)(mole moleNi(CO)4/moleNi) = 25.0 moleNi(CO)4
Therefore Ni is the limiting reagent, 25.0 moleNi(CO)4 are produced and excess CO = (5600gCO)(moleCO/28.0g) –(25.0 moleNi(CO)4)(4moleCO/moleNi(CO)4) =
100 mole CO left in excess mixed with the 25.0 moleNi(CO)4 thus
(25.0 moleNi(CO)4)/[( 25.0 moleNi(CO)4 + 100.0mole CO)=
= 0.200 moleNi(CO)4/mole total
7. 2 FeTiO3 + 7Cl2(g) + 3C → 2TiCl4 + 2 FeCl3 + 3CO2(g)
TiCl4 + 2Mg → Ti + 2MgCl2
a. (5000gFeTiO3)(mole FeTiO3/151.75g)(moleTiCl4/mole FeTiO3)(moleTi/mole TiCl4)(47.88gTi/mole) = 1.58x103 g Ti
b. (5000gFeTiO3)(mole FeTiO3/151.75g)(3moleCO2/2mole FeTiO3)(.08205Latm/
moleK)(296.15K/550mmHg)(760mmHg/atm) = 1.66x103 L CO2
8. MnO2 + 2NaCl + 2 H2SO4 → MnSO4 + Cl2(g) + 2H2O + Na2SO4
a. (1.00g MnO2)(moleMnO2/86.9g)(moleCl2/moleMnO2)=1.15x10-2 moleCl2
(1.00gNaCl)(moleNaCl/58.4g)(moleCl2/2moleNaCl) = 8.56x10-3 mole Cl2
(5.00g H2SO4)(mole H2SO4/98.1g H2SO4)(moleCl2/2mole H2SO4) =
2.55x10-2 mole Cl2
Thus NaCl is the limiting reactant and
(8.56x10-3 mole Cl2)(.08205Latm/moleK)(298.15K/620.0mmHg)(760mmHg/atm) = 0.260L or 260mL Cl2 is formed
b. (1.00L Cl2)(620mmHg)(atm/760mmHg)/[(.08205Latm/moleK)(298.15K)] = 3.33x10-2 mole Cl2
(.0333moleCl2)(2moleH2SO4/moleCl2)(98.1gH2SO4/mole)(g soln/.95gH2SO4)
(ml soln/1.84g soln) = 3.74 mL soln
9. (.200gCuFeS2/g ore)(mole CuFeS2/183.5g)(moleCu/moleCuFeS2)(63.5gCu/mole) = 0.0692 gCu/g ore
(.160gCu2(OH)2CO3/g ore)(mole Cu2(OH)2CO3/221g)(2moleCu/mole Cu2(OH)2CO3) (63.5gCu/mole) = .0919gCu/g ore or 9.19% Cu
10. a. (.0486L Fe3+/.1000L MnO4-)(eq MnO4-/eq Fe3+)(2.50eq Fe3+/L Fe3+) = 1.22 N
b. (1.22eq MnO4-/L)(mole MnO4-/5eq) = 0.243 M
c. (.243mole KMnO4/L soln)(158.03gKMnO4/mole)= 38.4 gKMnO4/L soln
1350g soln/L soln – 38.4 gKMnO4/L soln 1312gH2O/L soln
(.243mole KMnO4/L soln)(L soln/1312gH2O)(1000g/kgH2O) =
0.183 mole KMnO4/kgH2O or 0.185m
11. a. (.180gNH3/g soln)(.9295g soln/ml)(1000ml/L soln)(moleNH3/17.0g) = 9.84M
b. (.180gNH3/.820gH2O)(moleNH3/17.0g)(1000g/kgH2O) = 12.9 m
c. total moles = (.180gNH3)(mole/17.0gNH3) + (.820gH2O)(mole/18.0g) = .0561
(.180gNH3)(mole/17.0gNH3)/.0561 total = .189 moleNH3/total moles
12. 2C + 4H2 + O2 → 2CH3OH
(12.00gO2)(moleO2/32.0g)(2mole CH3OH/moleO2) = 0.75 mole CH3OH
(6.00gC)(moleC/12.0g)(mole CH3OH/moleC) = 0.500 mole CH3OH
So C is the limiting reactant and .500 mole CH3OH form
(.500moleCH3OH)(4moleH2/2moleCH3OH)(.08205Latm/moleK)(303K)/
(685torr)(atm/760torr) = 27.6 L H2
13. (4.00moleN2)(28.0g/mole) + 19.00 gF2 + (1.204x1024atomAr)(moleAr/6.02x1023 atom)(39.95g/mole) = 211 g total
4.00moleN2 + (19.00 gF2)(mole/38.0g) + (1.204x1024atomAr)(moleAr/6.02x1023
atom) = 6.50 mole total
a. (211g/6.50mole) = 32.4 g/mole
b. (625torr)(atm/760torr)/(.08205Latm/moleK)(323K) = 0.0310 mole/L
(0.0310 mole/L)(32.4g/mole) = 1.01 g/L
c. (19.00 gF2)(moleF2/38.0g)/6.50mole total = 0.0769 moleF2/total moles
14. a. (19.62gH2SO4/.8000L soln)(moleH2SO4/98.10g) = 0.2500 M
b. (800.0ml soln)(1.0115g soln/ml) – 19.62 g H2SO4 = 789.6 gH2O
(19.62gH2SO4)(mole/98.10g) + (789.6gH2O)(mole/18.0g) = 44.07 moles
(19.62gH2SO4)(mole/98.10g)/ 44.07 moles total = 4.538x10-3
c. (.1000L)(.2000mole/L)(L soln/.2500mole) =.0800L or 80.0mL soln
d. (.1000L)(.2500mole/L) = .02500 moles
15. (29.0gCsCl)(moleCsCl/168.35g)(L soln/6.00moleCsCl) = .02871L soln
(29.0+21.0 g soln)/(.02871L soln)(1000ml/L) = 1.75 g/ml
16. [(.0350L)(2.50mole/L) + (.0500L)(.150mole/L)]/.0850L = 1.12 mole/L
17. total moles = V(.250mole/L) + (.155L)(2.00mole/L)
total volume = V + .155 L
1.50mole/L = (.250V + .310)/(V + .155) V = .0620L or 62 mL
18. HCl + NaOH → NaCl + H2O
(.0300L HCl)(.125moleHCl/L)(moleNaOH/moleHCl)(LNaOH/.150moleNaOH) =
=.0250L NaOH or 25.0 mL
19. KHPhth + base → K+ + Phth2- + baseH+ + H2O
(.8650g sample)(.9995gKHPhth/g sample) = .8646 gKHPhth
(.8646gKHPhth/.04385L base)(moleKHPhth/204.2g)(molebase/moleKHPhth) =
= .09655 M base
20. (.02365L HCl)(0.1120moleHCl/L)(moleNa2CO3/2moleHCl)(106.0gNa2CO3/
mole) = 0.1404 gNa2CO3
(0.1404 gNa2CO3)/(0.485g sample) = 0.289 gNa2CO3)/g sample or 28.9%
21. NaOH + HA → NaA + H2O
(.7515gHA/.04320LNaOH)(LNaOH/.0904moleNaOH)(moleNaOH/moleHA) =
192.4 gHA/moleHA
22. OH- + H3O+ → 2H2O (HNO3 reacts with water to form H3O+ and NO3- and
NaOH in water consists of Na+ and Cl- ions)
(.02500LNaOH)(.1500moleOH-/L) = .00375 mole OH-
(.01500LHNO3)(.2000mole H3O+/L) = .00300 mole H3O+
So after reaction .00075 mole OH- remain, all H3O+ has been consumed
(.00075 moleOH-/.04000L total) = .01875 M OH-
(.02500LNaOH)(.1500moleNa+/L)/(.04000L total) = 0.09375 M Na+
(.01500LHNO3)(.2000moleNO3-/L)/ (.04000L total) = 0.07500 M NO3-
23. NH3 + HCl → NH4+ + Cl-
(.02500L NH3)(.1500moleNH3/L) = .00375 mole NH3
(.01000LHCl)(.1200moleHCl/L) = .001200 moleHCl
So after reaction .00375-.00120 = .00255 mole NH3 remain and .001200 mole NH4+ have formed using all the HCl
(.00255moleNH3/.03500L total) = 0.07286 M NH3
(.001200moleNH4+/.03500L) = 0.03429 M NH4+
24. HCl + NaOH → H2O + Na+ + Cl-
(.0550LHCl)(.200moleHCl/L) – (.0450LNaOH)(.150moleNaOH/L) (moleHCl/moleNaOH) = .00425 mole HCl left (in the form of H+ or H3O+ and Cl-)
So [H3O+] = .00425mole/.1000L total = .0425M
[Cl-] = (.0550L)(.200moleCl-/L)/(.1000L total) = 0.110 M Cl-
25. 2MnO4- + 10I- + 16 H3O+ → 2Mn2+ + 5I2 + 24H2O
(1.25gKI)(moleKI/166.0g)(2moleKMnO4/10moleKI)(L KMnO4/0.150mole) =
= 0.0100 L or 10.0 mL KMnO4
26. MnO4- + 5 Fe2+ + 8 H3O+ → Mn2+ + 5 Fe3+ + 12H2O
(.0500L Fe2+)(.100mole Fe2+/L)(mole MnO4-/5mole Fe2+)(L KMnO4/.150mole MnO4-) = 0.00667L or 6.67 mL KMnO4
27. 2 S2O32- + I2 → S4O62- + 2 I-
(0.300gI2)(moleI2/253.9g)(2mole S2O32-/moleI2)(LNa2S2O3/.200mole S2O32-) =
= 0.0118 L or 11.8 mL Na2S2O3
28. MnO2 + HAsO2 + 2 H3O+ → Mn2+ + H3AsO4 + 2H2O
(.0287L)(.0600moleHAsO2/L)(moleMnO2/moleHAsO2)(86.9gMnO2/mole) =
= 0.149 g MnO2
(0.149 g MnO2)/(0.200g sample) =0.749 g MnO2/g sample or 74.9% MnO2
29. a. -2 +1 +7 -2 +5 -2 +4 -2
N2H4 + MnO4- → NO3- + MnO2
(+14) (-3)
3N2H4 + 14MnO4- → 6NO3- + 14MnO2
3N2H4 + 14MnO4- → 6NO3- + 14MnO2 + 8 OH-
3N2H4 + 14MnO4- → 6NO3- + 14MnO2 + 8 OH- + 2H2O
b. +3 -1 +1 -1 0 0 -1
AuCl63- + H2O2 → O2 + Au + Cl-
(-3) (+2)
2AuCl63- + 3H2O2 → 3O2 + 2Au + 12Cl-
2AuCl63- + 3H2O2 → 3O2 + 2Au + 12Cl- + 6H3O+
2AuCl63- + 3H2O2 + 6H2O → 3O2 + 2Au + 12Cl- + 6H3O+
c. +3 -2 +4 +1 +5 -2 +4 -2 +3
As2S3 + Ce4+ → H3AsO4 + SO2 + Ce3+
(+22) (-1)
As2S3 + 22Ce4+ → 2H3AsO4 + 3SO2 + 22Ce3+
As2S3 + 22Ce4+ → 2H3AsO4 + 3SO2 + 22Ce3+ + 22H3O+
As2S3 + 22Ce4+ + 36H2O → 2H3AsO4 + 3SO2 + 22Ce3+ + 22H3O+
d. +1 -2 +5 -2 +2 +2 -2 0
Cu2S + NO3- → Cu2+ + NO + S
(+4) (-3)
3Cu2S + 4NO3- → 6Cu2+ + 4NO + 3S
3Cu2S + 4NO3- + 16H3O+ → 6Cu2+ + 4NO + 3S
3Cu2S + 4NO3- + 16H3O+ → 6Cu2+ + 4NO + 3S + 24H2O
30. (5.60gN2)(moleN2/28.02g) =0.200 mole N2
(7.20gO2)(moleO2/32.0g) = 0.225 moleO2
a. PN2 = (700torr)(.200 moleN2/.425 mole total) = 329torr
b. (12.80 mole total/.425 mole total) = 30.1 g/mole
c. V = nRT/P = (.425mole)(.08205Latm/moleK)(313.15K)/(700/760 atm) =
= 11.9 L
31. (3.08gN2)(mole/28.0gN2) = 0.110 mole N2
(2.85gF2)(mole/38.0gF2) = .0750 mole F2
(.110 mole N2)(2 mole NF3/mole N2) = .220 mole NF3 but
(.0750 mole F2)(2 mole NF3/3 mole F2) = .0500 mole NF3 so F2 is limiting
.110 mole N2 – (.0500 mole NF3)( mole N2/2 mole NF3) = .0850 mole N2 left
Total moles = .0850 + .0500 = .135 moles
a. V= nRT/P = (.135 mole)(.08205Latm/moleK)(343K)/(1.30atm) = 2.92 L
b. Initially total moles = .110 +.0750 = .185moles
P= (.185 mole)(.08205Latm/moleK)(343K)/(2.92L) = 1.78 atm
c. (.0850 mole N2)/(.135 mole total) = .630 mole fraction N2
d. (.0500 mole NF3)/(.135 mole total) = .370 mole fraction NF3
32. Total mass is constant.
(2.00mole H2)(1.008g/mole) + (2.00mole S2)(64.12g/mole) = 132.3 g total
(132.3g)/(.534g/L) = 247.7 L volume
n = PV/RT = (2.00atm)(247.7L)/(.08205Latm/moleK)(1573K) = 3.84 moles
2 H2 + S2 ↔ 2 H2S
2.00-2x 2.00-x 2x so total moles = 2.00-2x+2.00-x+2x = 4.00-x
a. 4.00-x = 3.84 therefore x= 0.16 and moles H2S = 0.34 moles H2S
b. PH2S = (0.34 mole)(.08205Latm/moleK)(1573K)/(247.7L) =0.18 atm
PH2 = (1.66mole) )(.08205Latm/moleK)(1573K)/(247.7L) = 0.87 atm
PS2 = (1.84mole)RT/V = 0.95 atm
33. (60.0gCu)(80.0oC-T)(0.0924cal/goC) = (20.0gH2O)(T-10.0oC)(1.00cal/goC)
T = 25.2oC
34. (95.0g ice)(1435cal/mole)(mole/18.0g ice) = 7.57x104 cal to melt all ice
(155gFe)(215-0oC)(0.106calgFeoC) = 3.53x104 cal to cool Fe to 0.0oC
Therefore the Fe is cooled to 0.0oC melting only part of the ice to water
(3.53x104 cal)(mole/1435cal)(18.0g ice/mole) = 44.3g ice melt
44.3 g water + 50.7g ice at final temperature of 0.0o
35. 2C(gr) + 2O2(g) → 2CO2(g) ∆Ho1
3H2(g) + 3/2 O2(g) → 3H2O(l) ∆Ho2
2CO2(g) + 3H2O(l) → C2H6(g) ∆Ho3
2C(gr) + 3H2(g) → C2H6(g) ∆Hof = ∆Ho1 + ∆Ho2 + ∆Ho3
∆Ho1 = (2moleC)(-94.05kcal/mole) = -188.10 kcal
∆Ho2 = (3moleH2)(-68.32kcal/mole) = -204.96 kcal
∆Ho3 = (moleC2H6)(+372.8kcal/mole) = +372.8 kcal
∆Hof = -188.10 -204.96 + 372.8 = -20.3 kcal/moleC2H6(g)
36. a. HCl and HF both acidic, Cl- is very weak conjugate base of the very strong acid HCl so NaCl is neutral, but F- is the weak conjugate base of the weak acid HF. Thus NaF is basic and has the highest pH.
b. Addition of the weak base NaF to HF raises the pH while the much weaker base NaCl does not significantly raise the pH when added to HCl. Therefore the HF/NaF solution has the highest pH.
c. NH4+ is the weak conjugate acid of the weak base NH3. Addition of NH4Cl to basic NH3 therefore lowers the pH. So the NH3 solution has the highest pH.
d. NaCl is neutral because is Cl- such a weak conjugate base. NH4Cl is acidic because NH4+ is a weak acid and Cl- a very weak base. NH4F is slightly acidic because NH4+ is stronger acid than is F- a base. So NaCl has the highest pH.
e. NaOH is a strong base and HF is a weak acid so equivalence point pH is above 7; NaOH and HCl are both strong so eq. pt. pH is 7; NH3 is a weak base and HCl a strong acid so eq. pyt. pH is acidic. So NaOH/HF has the highest eq. pt. pH.
37. (.0500L HF)(0.100moleHF/L)(moleNaOH/moleHF)(L NaOH/0.400moleNaOH) = 0.0125 L NaOH at the equivalence point. Total V = 0.0625L
HF + NaOH → NaF + H2O so at eq. pt. solution is NaF solution of concentration
(.0500L HF)(.100moleHF/L)(moleNaF/moleHF)/.0625L = 0.0800 M HF
F- + H2O ↔ HF + OH- K = [HF][OH-]/[F-] x [H3O+]/[H3O+] = Kw/Ka
F- + H2O ↔ HF + OH- K = 1.0x10-14/4.8x10-4 = 2.1x10-11
.080-x x x
(x)(x)/(.080-x) = 2.1x10-11 x= 1.3x10-6 = [OH-]
pOH = -log[OH-] = 5.89 pH = 14.00-pOH = 8.11
38. B + H3O+ → BH+ + H2O
(.0500L B)(.300mole B/L) = .0150 moleB
(.0250L HCl)(.200mole H3O+/L) = 0.00500 mole H3O+
So after reaction .0150mole B – (.00500mole H3O+)(moleB/ mole H3O+) =
= .0100 mole B left and .00500 mole BH+ formed in .0750 L total
B + H2O ↔ BH+ + OH- Kb = [BH+][ OH-]/[B]
(.0100/.075) -x (.00500/.0750) + x x
But pH = 8.25, so pOH = 5.75 and [OH-] = 10-pOH = 1.78x10-6 = x
Kb = [BH+][ OH-]/[B] = (.0667+1.78x10-6)( 1.78x10-6)/(.133-1.78x10-6) =
= 8.9x10-7
39. a. Greater NaCN is the conjugate base of the weak acid HCN
b. Greater NaCl is the weak conjugate base of the strong acid HCL while NaCN is the stronger base of the weak acid HCN.
c. Greater The more concentrated base has the higher pH
d. Less NH4NO3 contains the fairly strong conjugate acid NH4+ of NH3 and the very weak conjugate base NO3- of the strong acid HNO3 so it is acidic and lowers the pH.
e. Greater due to the basicity of NH3.
f. Less The strong acid HCl wins out over an equal amount of the weak base.
g. Same The resulting mixture at the equivalence point is simply 0.500M NH4Cl.
h. Greater The less concentrated solution of the weak conjugate acid has a higher pH.
i. Less HCl is a stronger acid than is HF.
40. a. HSCN + H2O ↔ H3O+ + SCN- Ka = [H3O+][SCN-]/[HSCN] = 0.14
.100-x x x
0.14 = (x)(x)/(.100-x) x= 0.067 pH = -log[H3O+] =-logx = 1.17
b. HSCN + OH- → H2O + SCN-
(.0500L HSCN)(.100mole/L) – (.0150LNaOH)(.200moleNaOH/L) (moleHSCN/moleNaOH) = 0.0020 mole HSCN left in .0650L total
(.0150LNaOH)(.200moleNaOH/L)(moleSCN-/moleNaOH) =
0.0030 mole SCN- formed
HSCN + H2O ↔ H3O+ + SCN-
.0020/.0650)-x x (.00300/.0650)+x
0.14 = (x)(.0462+x)/(.0308-x) x = 0.021 pH = -logx = 1.68
41. a. Electronegativity generally increases left to right and bottom to top in the Periodic Table. Therefore S has the greatest electronegativity among these.
b. Ionization potential increases left to right and bottom to top in the Periodic Table. Ionization of a second electron requires a higher potential than that for the first electron. Therefore K has the small ionization potential among these.
c. Ar has a higher ionization potential than S or Ca as explained above. Removal of an electron from S2- is easer than from S. Ca2+ and Ar contain the same number of electrons but there are 2 more nuclear protons in Ca2+ which therefore attract the electrons more strongly. So Ca2+ has the highest ionization potential among these.
d. Atomic size decreases left to right and bottom to top in the Periodic Table. Anions are larger than the corresponding atoms while cations are smaller. Thus I- has the largest radius among these.
42. a. 2 Hg + 2 Cl- → Hg2Cl2 + 2 e- anode reaction
IO3- + 6 H3O+ + 6 e- → I- + 9 H2O cathode reaction
IO3- + 6 H3O+ + 6 Hg + 6 Cl- → 3 Hg2Cl2 + I- + 9 H2O cell reaction
b. Eocell = Eanode + Ecathode = -.355 +1.085 = +.730 volt
c. Ecell = Eocell –(.0591/n)log([I-]/[ Cl-]6[IO3-][ H3O+]6)
= +.730-(.0591/6)log{(.500)/(.100)6(5.00)(1.0x10-3)6} = +.503 volt
d. (.600coul/sec)(3.00hr)(3600sec/hr)(eq/96500coul)(moleI-/eq) =
= 0.0112 mole I- produced
[I-] = .500 + (.0112mole/.200L) = 0.556 M
43. HBrO + 7 H2O → BrO3- + 4 e- + 5 H3O+ ∆Go = -(4eq)(96500coul/eq)(-1.59v)
BrO3- + 6 H3O+ + 5 e- → ½ Br2 + 9 H2O ∆Go = -(5eq)(96500coul/eq)(+1.52 v)
HBrO + H3O+ + e- → ½ Br2 + 2 H2O ∆Go = -119660 v-coulomb
Eo = -∆Go/nF = -(-119660 v-coulomb)/(1)(96500coulomb) = +1.24 v.
44. AgBr(s) + e- → Ag(s) + Br- ∆Go = -(1eq)(96500coul/eq)(+.071v)
Ag → Ag+ + e- ∆Go = -(1eq)(96500coul/eq)(-.80v)
AgBr(s) → Ag+ + + Br- ∆Go =.73(96500) = +7.04x104 J
Ksp = exp{-∆Go/RT} = 4.5x10-13
45. 8 OH- + S2O42- → 2 SO42- + 6 e- + 4 H2O ∆Go = -6F(+.99)
2 H2O + 4 e- + 2 SO42- → 2 SO32- + 4 OH- ∆Go = -4F(-.93)
4 OH- + S2O42- → 2 SO32- + 2 H2O + 2 e- ∆Go = -2.22F
Eo = -(-2.22F)/2F = +1.11 volt
46. PbSO4(s) → Pb2+ + SO42- ∆Go =-RTlnKsp = -(8.314)(2998.15)ln(1.6x10-8)
Pb2+ + 2 e- → Pb ∆Go = -nFEo = -(2)(96500)(-.126)
PbSO4(s) + 2 e- → Pb + SO42- ∆Go =+4.5x104 J + 2.4x104 J = +6.88x104 J
Eo = -6.88x104/(2)(96500) = -.36 volt
47. a. Cd2+ + 2 e- → Cd Eo = -.402 v.
2 Cl- → Cl2(g) + 2 e- Eo = -1.3595 v.
Cd2+ + 2 Cl- → Cl2(g) + Cd Eo = -1.761 v.
E = -1.761 –(.0591/2)log{(1.00atm)/(3.00)(6.00)2} =-1.701 v.
So decomposition potential is +1.701 v.
b. 2 Cl- → Cl2(g) + 2 e- is the anode reaction
c. + pole
48. There are two possible anode reactions:
2 Cl- → Cl2 + 2e- E= E0 = -1.3595v since [Cl-] = 1.00M and Cl2 gas presumably is at 1 atm pressure
2 H2O → O2 + 4 H+ + 4 e- E = -1.229 – (.05917/4)log[H+]4PO2
Assuming pH is 7.00 and oxygen gas is at 1.00 Atm, then
E = -1.229 – (.05917/4) log(1.0x10-7)4(1.00 atm) = -0.815 v.
So oxidation of water is the anode reaction. The cathode reaction is
Ni2+ + 2 e- → Ni E= -.250 –(.05917/2)log(1/0.500) = -0.258v
a. So decomposition potential is ).815+0.258 = 1.074 v.
b. 2 (Ni2+ + 2 e- → Ni) cathode reaction
2 H2O → O2 + 4 H+ + 4 e- anode reaction
2 H2O + 2 Ni2+ → O2 + 4 H+ + 2 Ni cell reaction
c. Anode as electrodes must be removed from the anode to produce oxidation.
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- solutions worksheet 1 solutions electrolyte s and
- solutions to practice problems
- colligative properties
- ap chemistry study guide for solutions chapter 11
- chapter 5 homework problems chemistry
- molarity practice worksheet
- exercises mass percent the chemistry geek s refuge on
- chemistry worksheet wavelength frequency energy of
Related searches
- sat math practice problems printable
- significant figures practice problems worksheet
- theoretical yield practice problems answers
- practice problems for significant figures
- synthesis practice problems and answers
- sig fig practice problems worksheet
- significant figures practice problems pdf
- percent yield practice problems answers
- statistic practice problems with answers
- organic chemistry practice problems pdf
- percent yield practice problems worksheet
- practice problems for percent yield