2019 Final Exam

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University of British Columbia *

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358

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Material Science

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Jan 9, 2024

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THE UNIVERSITY OF BRITISH COLUMBIA Department of Materials Engineering MTRL 358 Hydrometallurgy I Final Examination 2019 [Total marks = 54] Total time: 2.5 hours Answer all questions This examination has 11 pages (including supplementary material). Formulas and data are included on pages 9-11. Marks for each question are shown in square brackets. 1. [6 marks] Briefly answer the following questions: (a) What are the two purposes of solvent extraction? Purification and upgrading of the PLS (upgrading means to increase the concentration of the desired species). (b) Indicate whether the following sentence is true or false: Leaching is the process of dissolving a solid, in whole or in part, into an aqueous solution. TRUE (c) Name two hydrometallurgical process steps or operations in which the principle of counter-current flows are applied. Any two of counter-current leaching, counter-current decantation and counter-current solvent extraction (just solvent extraction is acceptable). (d) What happens at the cathode in an electrochemical cell? Reduction , or electron transfer to effect reduction. (e) What is the name for the aqueous stream exiting a solvent extraction loading process? Raffinate This exam is closed book and closed notes. Read all questions carefully before beginning and work neatly. Calculators are allowed. Cell phones and other electronic devices MUST be left at the front of the exam room. Backpacks, bags, coats etc. MUST be left at the front of the exam room.
(f) Silver sulfide (Ag 2 S) is highly insoluble in water (K sp = 6 x 10 -51 ). True or false: because of its low solubility it can’t be leached. False 2. [5 marks] The mineral enargite has the formula Cu 3 AsS 4 . Write a balanced chemical reaction for leaching of enargite by oxygen in sulfuric acid to form cupric ion, sulfate and arsenic acid (H 3 AsO 4 ; a soluble weak acid assume it is undissociated). Write the reaction in ionic and neutral forms. Include the phases in the reaction in neutral form. 2Cu 3 AsS 4 + 5H 2 O + 17.5O 2 = 6Cu +2 + 2H 3 AsO 4 + 8SO 4 2- + 4H + 2Cu 3 AsS 4 s + 5H 2 O l + 17.5O 2 g = 6CuSO 4 aq + 2H 3 AsO 4 aq + 2H 2 SO 4 aq 3. [5 marks] The mineral CuS reacts with cyanide according to the following reaction: 2CuS + 8CN - + 0.5O 2 + H 2 O = 2[Cu(CN) 3 ] 2- + 2SCN - + 2OH - An ore contains 0.11% copper (1.1 kg Cu/tonne of ore) in the form of CuS. The ore is processed at a rate of 1500 tonnes per hour. Assuming 75% of the mineral is leached, how much sodium cyanide would be consumed by this reaction in kg/h? 3820 kg NaCN/h 4. [8 marks] An Eh-pH diagram for the Fe-S-H 2 O system involving only pyrite is shown below. (Pyrite itself is not valuable, but it may contain gold.) Conditions are indicated in the diagram. Thermodynamic data are also shown in the table below. Species f kJ Species f kJ Fe +2 -78.7 H 2 S g -33.56 Fe(OH) 2 -492 HS - 11.44 Fe +3 -4.6 S 8 ( or S) 0 Fe(OH) 3 -704.63 SO 4 2- -744.55 FeS 2 -166 HSO 4 - -755.91 H 2 O l -237.15 (a) Is it possible to leach iron from FeS 2 at pH 8? Briefly explain your answer. No. At pH 8 reduction of FeS 2 forms Fe s and HS - . There is no soluble Fe species. Oxidation forms solid Fe(OH) 3 . Either way Fe is not dissolved.
(b) Write the balanced half reaction for the line numbered (1) in the diagram. Fe +2 + 2SO 4 2- + 16H + + 14e - = FeS 2 + 8H 2 O (c) Calculate the equation for the line in the form Eh = Slope x pH + Constant. Eh = -0.06762pH + 0.36125 (d) (i) Write the main reactants and products for the two half reactions for leaching of pyrite by ferric ion at pH 0.5. (The balanced half reactions are not necessary, just the main reactants and products for each.) Fe +3 /Fe +2 Fe +2 ,HSO 4 - /FeS 2 (ii) Estimate the Eh values (including the units) for these two half reactions from the diagram. Eh Fe+3/Fe+2 = 0.77 V; Eh Fe+2/HSO4-/FeS2 = 0.32 V (iii) Estimate the E for the reaction. E =0.45 V (iv) Is the reaction favourable? Briefly explain your answer. Yes; E > 0 -1.0 -0.5 0.0 0.5 1.0 0 2 4 6 8 10 12 Eh V pH Eh-pH diagram for the iron-sulfur-water system at 25°C and 1 atm pressure, 0.2 m Fe solute activities and 0.5 m S solute activities. Fe +3 Fe +2 HSO 4 - SO 4 2- Fe(OH) 3 Fe +2 Fe Fe(OH) 2 H 2 S g HS - FeS 2 H+ H2 O2 H2O 1
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5. [5 marks] Below is a simplified flowsheet for a zinc roast-leach-electrowin process. Zinc calcine is obtained by roasting a zinc sulfide concentrate which contains ZnS, FeS 2 and small amounts of other minerals. The calcine contains mainly ZnO and zinc ferrite (ZnFe 2 O 4 ). Electrowinning produces most of the sulfuric acid needed in leaching. The jarosite precipitation stage forms solid NH 4 Fe 3 (SO 4 ) 2 (OH) 6 , which is highly insoluble. Suitable conditions for jarosite precipitation are about 95°C and pH ~1.5. (This is not sufficient to leach the zinc ferrite.) A small fraction of the zinc calcine is used to adjust pH in jarosite precipitation. Jarosite will not redissolve in the hot acid leach. (a) In which step does Fe(III) enter solution? The hot acid leach (b) Is the jarosite precipitation step primarily a leaching process or a solution purification process? Briefly explain your answer. Solution purification. Fe(III) is removed from solution as jarosite which is separated from solution by solid-liquid separation. (Some leaching of calcine occurs in that it acts as a base for neutralization of some of the acid, but that is not the main point.) (c) Indicate which of the following statements are true: (i) This is an example of a co-current leaching process. (ii) A disadvantage of this process is that zinc in the form of zinc ferrite added with the calcine into jarosite precipitation is not recovered. (iii) In this process zinc present in the calcine as zinc ferrite is leached in the hot acid leach step. True (iv) In this process zinc present in the calcine as zinc ferrite is leached in the neutral leach step. (v) Good use of sulfuric acid in leaching is ensured by directing it back towards the neutral leach. True (vi) Zinc oxide is mainly dissolved in the neutral leach step. True (vii) Zinc oxide is mainly dissolved in the hot acid leach step.
6. [8 marks] A gold ore is finely ground and then leached. The leach solution is to be separated from the solids using CCD. The leach slurry contains 31% solids and the solids flow rate into the process is 930 tonnes/hour. The leach solution gold concentration is 1.8 mg/L. The wash ratio is 1.8. The underflow slurry from each thickener contains 40% solids. There are a total of 6 thickeners in the circuit. The solutions have a density of about 1 t/m 3 . (a) Calculate the feed solution flow rate in m 3 /h. 2070 m 3 /h (b) Calculate the underflow solution flow rate and the wash water flow rate in m 3 /h. 1395 m 3 /h; 2511 m 3 /h, respectively Neutral leach Zinc calcine L S Jarosite precipitation L S Hot acid leach L S Residues (NH 4 ) 2 SO 4 Zinc calcine Zinc dust cementation Electrowinning Zinc metal Spent electrolyte Spent electrolyte Extra H 2 SO 4 L S Residues
(c) Calculate the clear overflow solution flow rate from the lead thickener in m 3 /h. 3186 m 3 /h (d) How many wash stages are there in the circuit? 5 (e) Calculate the wash efficiency and the recovery of gold in % in the clear overflow solution from the lead thickener. 97.6%, 99.0%, respectively (f) Calculate the gold concentration in the clear overflow solution from the lead thickener in mg/L. 1.16 mg/L 7. [8 marks] A copper SX plant has a 2E 1S configuration. The stripping isotherm for copper using 7% Lix 984N and a lean electrolyte [Cu +2 ] of 35 g/L is shown below (next page). (a) The rich electrolyte contains 47 g/L Cu +2 . Estimate the [Cu +2 ] in the barren organic. From the stripping isotherm: 1.40 g/L (See diagram) (b) The PLS contains 1.8 g/L Cu +2 and has a pH of 1.8. The copper extraction from the PLS is 90%. What is the raffinate copper concentration in g/L? 0.18 g/L (c) The loaded organic copper concentration is 3.3 g/L. What is the A/O ratio? Show your calculation. A/O = y/ x = (3.3-1.40)/(1.8 0.18) = 1.173 (d) Two copies of a loading curve are provided below (pages 7 and 8). Draw a McCabe-Thiele diagram for the solvent extraction loading process. Indicate the stage numbers on the diagram. Indicate the 4 streams involved, e.g. the PLS, etc. Detach your final diagram from the exam paper and include it with your answer booklet. Make sure your name is on the diagram. See diagram below (e) If the copper production rate is 50,000 tonnes per year what is the electrolyte flow rate in m 3 /h? (Assume 365 days per year operation; no down time.) 475.6 m 3 /h
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Stripping isotherm for Question 7: 30 35 40 45 50 55 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Electrolyte [Cu +2 ] g/L Organic [Cu +2 ] g/L Stripping isotherm for 7% lix 984N with 180 g/L H 2 SO 4 , 35 g/L Cu +2
0 1 2 3 4 0 0.5 1 1.5 2 2.5 Organic [Cu +2 ] g/L Aqueous [Cu +2 ] g/L Loading curve for 7% Lix 984N Name: Barren organic Loaded organic PLS Raffinate 1 2 Question 7 McCabe-Thiele diagram
8. [9 marks] A copper electrowinning process consists of 172 cells, each containing 60 cathodes. The cathode immersed dimensions are 1.1 m x 1 m. The current efficiency is 90% and the energy consumption is 1940 kWh/tonne Cu. The annual production rate is 64,300 tonnes of copper. (Assume 365 days/year operation.) (a) What is the applied voltage, including the units? 2.07 V (b) What is the current density in A/m 2 ? 303 A/m 2 (c) In a different plant the current efficiency is 91% and the current density is 247 A/m 2 . The immersed cathode dimensions are 1.1 m x 1.0 m. The density of copper is 8.94 g/cm 3 . How long ( in days ) will it take to plate the cathodes to a thickness of 0.50 cm? 6.99 days (d) What is the bare minimum applied cell voltage required for copper electrowinning? 0.89 V (positive an applied voltage) (e) Explain why the actual applied cell voltage is much greater than the minimum cell voltage. In addition to overcoming the thermodynamic voltage (- E) extra driving force is needed to force the oxygen evolution (anodic) half reaction ( O2 ) and the copper reduction (cathodic) half reaction ( Cu ). There are also voltage drops due to resistances such as the electrical conductors and the solution (I R i ). In total these make E appl significantly greater than E.
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Formulas and data R = F/U + x 1 where x = O/U F/U + x - x n+1 x x n+1 1 For a generalized half reaction: aA + bB + mH + + ne - = cC + dD Eh = - - 2.303RT log a C c a D d - 2.303RTm pH nF nF a A a a B b nF For a generalized hydrolysis reaction: aA+ bB + mH + = cC + dD pH = - - 1 log a C c a D d 2.303RTm m a A a a B b E = E° - 2.303RT logQ nF Faraday’s law relationships: q = n M nF M = It AW nF Constants: CE = 100nFM R = 8.3145 J/mol K = 0.08206 L atm/mol K ItAW EE =  E rev CE F = 96485 C/mol e - E appl -w e ’ = 100nF E appl Conversions: CE 1 kWh = 3.6 10 6 J dn M = jA c NCE dt 100nF 0°C = 273.15 K
2007 standard atomic weights Atomic Atomic Atomic weight weight weight actinium Ac 227 hafnium Hf 178.49 promethium Pm 145 aluminium Al 26.981538 hassium Hs 269 protactinium Pa 231.03588 americium Am 243 helium He 4.002602 radium Ra 226 antimony Sb 121.76 holmium Ho 164.93032 radon Rn 222 argon Ar 39.948 hydrogen H 1.00794 rhenium Re 186.207 arsenic As 74.9216 indium In 114.818 rhodium Rh 102.9055 astatine At 210 iodine I 126.90447 rubidium Rb 85.4678 barium Ba 137.327 iridium Ir 192.217 ruthenium Ru 101.07 berkelium Bk 247 iron Fe 55.845 rutherfordium Rf 261 beryllium Be 9.012182 krypton Kr 83.8 samarium Sm 150.36 bismuth Bi 208.98038 lanthanum La 138.9055 scandium Sc 44.95591 bohrium Bh 264 lawrencium Lr 262 seaborgium Sg 266 boron B 10.811 lead Pb 207.2 selenium Se 78.96 bromine Br 79.904 lithium Li 6.941 silicon Si 28.0855 cadmium Cd 112.411 lutetium Lu 174.9668 silver Ag 107.8682 caesium Cs 132.90545 magnesium Mg 24.305 sodium Na 22.98977 calcium Ca 40.078 manganese Mn 54.938049 strontium Sr 87.62 californium Cf 251 meitnerium Mt 268 sulphur S 32.066 carbon C 12.0107 mendelevium Md 258 tantalum Ta 180.9479 cerium Ce 140.116 mercury Hg 200.59 technetium Tc 98 chlorine Cl 35.4527 molybdenum Mo 95.96 tellurium Te 127.6 chromium Cr 51.9961 neodymium Nd 144.24 terbium Tb 158.92534 cobalt Co 58.9332 neon Ne 20.1797 thallium Tl 204.3833 copper Cu 63.546 neptunium Np 237 thorium Th 232.0381 curium Cm 247 nickel Ni 58.6934 thulium Tm 168.93421 dubnium Db 262 niobium Nb 92.90638 tin Sn 118.71 dysprosium Dy 162.5 nitrogen N 14.00674 titanium Ti 47.867 einsteinium Es 252 nobelium No 259 tungsten W 183.84 erbium Er 167.26 osmium Os 190.23 copernicium Cn 277 europium Eu 151.964 oxygen O 15.9994 darmstadtium Ds 269 fermium Fm 257 palladium Pd 106.42 roentgenium Rg 272 fluorine F 18.9984032 phosphorus P 30.973762 uranium U 238.0289 francium Fr 223 platinum Pt 195.078 vanadium V 50.9415 gadolinium Gd 157.25 plutonium Pu 244 xenon Xe 131.29 gallium Ga 69.723 polonium Po 210 ytterbium Yb 173.04 germanium Ge 72.61 potassium K 39.0983 yttrium Y 88.90585 gold Au 196.96655 praseodymium Pr 140.90765 zinc Zn 65.38 zirconium Zr 91.224 Name and symbol Name and symbol Name and symbol
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