2019 Final Exam
.pdf
keyboard_arrow_up
School
University of British Columbia *
*We aren’t endorsed by this school
Course
358
Subject
Material Science
Date
Jan 9, 2024
Type
Pages
12
Uploaded by noone00
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
G°
f
kJ
Species
G°
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
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
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help