Quiz Solutions

1. A simplified flowsheet for the Galvanox process is shown below. This process is being developed at
UBC. Chaclcopyrite is the most abundant mineral of copper in nature. However, it resists leaching by
conventional hydrometallurgical methods; the mineral rapidly passivates (after a short while it leaches
very slowly) when leached in sulfate solution. In this process a copper concentrate is produced which
contains pyrite and copper minerals, particularly chalcopyrite. If pyrite (FeS 2 ) is present in the
concentrate then chalcopyrite is readily leached. Pyrite, being an iron mineral, often occurs
in sulfide mineral deposits. During leaching
chalcopyrite is oxidized to form a solution
of Cu +2 , Fe +2 and solid elemental sulfur.
Pyrite itself undergoes little reaction.
(Mainly it seems to act as a catalyst for
oxygen reduction.) After leaching the slurry
is subjected to solid-liquid separation. The
solids may be recycled to leaching to
reutilize the pyrite. Much of the solution
proceeds to solvent extraction, which is
used to obtain a much purer copper solution.
The details are not important here. Some of
the solution also proceeds to an oxyhydrolysis
step (this is done in an autoclave) in which
ferrous ion is oxidized to form hematite (Fe 2 O 3 )
and sulfuric acid. This acts as an outlet for iron
and prevents its build-up in solution. Hematite
is a very suitable iron product for disposal.
The solvent extraction process also generates
acid, which together with that formed by
hematite formation can be reused in the
leaching step. The concentrated copper
sulfate solution from solvent extraction
proceeds to an electrowinning step. Here
very pure copper metal is produced by
electrolysis.

Faculty of Applied Sciences
Department of Materials Engineering

MTRL 358: Hydrometallurgy I
Assignment #2 - SOLUTIONS

Concentrate:
CuFeS
2, FeS
2

S
L
Iron
oxyhydrolysis

L
S

Solvent
extraction
Electrowinning
Copper metal

Solids
recycle
(FeS
2)

O
2

O
2

Fe
2O
3gangueH
2SO
4

Ore
gyratory
crusherOre

stockpileSAG
mill
cyclone
ball mill

Tailings

flotation

H
2SO
4
Concentrate:
CuFeS
2, FeS
2

S
L
S
L
Iron
oxyhydrolysis

L
S

Solvent
extraction
Electrowinning
Copper metal

Solids
recycle
(FeS
2)

O
2

O
2

Fe
2O
3gangueH
2SO
4

Ore
gyratory
crusherOre

stockpileSAG
mill
cyclone
ball mill

Tailings

flotation

H
2SO
4

Answer the following questions related to the flowsheet.
(a) Write the balanced chemical reaction in ionic and neutral forms for the leaching of chalocopyrite.
CuFeS 2 + O 2 = Cu +2 + Fe +2 + S
Oxygen is the oxidant, as seen from the flowsheet. The rest is evident from the text.
(a) CuFeS 2 = Cu +2 + Fe +2 + S O 2 = ?
(b) CuFeS 2 = Cu +2 + Fe +2 + 2S O 2 = ?
(c) CuFeS 2 = Cu +2 + Fe +2 + 2S O 2 = + 2H 2 O
(d) CuFeS 2 = Cu +2 + Fe +2 + 2S O 2 + 4H + = + 2H 2 O
(e) CuFeS 2 = Cu +2 + Fe +2 + 2S + 4e - O 2 + 4H + + 4e - = + 2H 2 O
(f) CuFeS 2 = Cu +2 + Fe +2 + 2S + 4e -
O 2 + 4H + + 4e - = + 2H 2 O
CuFeS 2 O 2 + 4H + = Cu +2 + Fe +2 + 2S + 2H 2 O ionic form
Balance + charges with sulfate:
CuFeS 2 s + O 2 g + 2H 2 SO 4 aq = CuSO 4 aq + FeSO 4 aq + 2S s + 2H 2 O l neutral form
(g) CHECK:
Left side Right side
1Cu 1Cu
1Fe 1Fe
4S 4S
10O 10O
4H 4H
0 charge 0 charge OK

(b) Referring to the generalized hydrometallurgical flowsheet on p. 30 of the Introduction to Extractive
Metallurgy course notes, identify the parts of the flowsheet that correspond to mineral separation,
leaching, solution purification and metal production. Use the flowsheet on the page below, circle and
label the appropriate parts of the flowsheet. Hand this in with your completed assignment.
See flowsheet below

(c) Referring to the generalized metallurgical flowsheet that follows, indicate which route corresponds
most closely to this process. You may indicate just the number in your assignment paper.
See flowsheet below
Copy of flowsheet; use this for question 1 (b).

Concentrate:
CuFeS
2, FeS
2

S
L

Iron
oxyhydrolysis

L
S

Solvent
extraction

Electrowinning

Copper metal

Solids
recycle
(FeS
2)

O
2

O
2

Fe
2O
3gangueH
2SO
4

Ore
gyratory
crusherOre

stockpileSAG
mill

cyclone
ball mill

Tailings

flotation

H
2SO
4

Concentrate:
CuFeS
2, FeS
2

S
L
S
L

Iron
oxyhydrolysis

L
S

Solvent
extraction

Electrowinning

Copper metal

Solids
recycle
(FeS
2)

O
2

O
2

Fe
2O
3gangueH
2SO
4

Ore
gyratory
crusherOre

stockpileSAG
mill

cyclone
ball mill

Tailings

flotation

H
2SO
4

Mineral separation -
size reduction
through to flotation

Leaching - mineral
dissolution

Solution purification-
iron is removed from
the solution.

Solution purification-
selective for copper

metal production - by
electrolysis

Generalized extractive metallurgy flowsheet
1 (c). Route 3 makes the most sense. A concentrate is made and then leached directly. The solution is
purified and then metal is precipitated by electrowinning.

Ore

Leach Gangue Particle size reduction

Ground ore

Gangue Leach Physical separation Gangue

Concentrate

Pyro-pretreatment

Gangue LeachTreated concentrate

SolutionLeach Gangue

Purification

Pure solution

Full pyro-processing to
yield matte or metal

Metal compound
precipitate

By-product Hydrometrefining
of metal

To market To Market To market To market? To market

Refining

Pure metal

Precipitated
metal

1

2

3

4

5

Ore

Leach Gangue Particle size reduction

Ground ore

Gangue Leach Physical separation Gangue

Concentrate

Pyro-pretreatment

Gangue LeachTreated concentrate

SolutionLeach Gangue

Purification

Pure solution

Full pyro-processing to
yield matte or metal

Metal compound
precipitate

By-product Hydrometrefining
of metal

To market To Market To market To market? To market

Refining

Pure metal

Precipitated
metal

11

22

33

44

55

1 (d). Leaching requires 4 moles of H + /mole of CuFeS 2 . What fraction of the required acid is produced by
the oxyhydrolysis step? (Consider the balanced reaction that forms Fe 2 O 3 .)
Here a chemical reaction is needed:
Fe +2 + O 2 = Fe 2 O 3
(a) Fe +2 = Fe 2 O 3 O 2 = ?
(b) 2Fe +2 = Fe 2 O 3
(c) 2Fe +2 + 3H 2 O = Fe 2 O 3
(d) 2Fe +2 + 3H 2 O = Fe 2 O 3 + 6H +
(e) 2Fe +2 + 3H 2 O = Fe 2 O 3 + 6H + + 2e - O 2 + 4H + + 4e - = 2H 2 O
(f) 2Fe +2 + 3H 2 O = Fe 2 O 3 + 6H + + 2e -
0.5O 2 + 2H + + 2e - = H 2 O
2Fe +2 + 2H 2 O + 0.5O 2 = Fe 2 O 3 + 4H + (Required to estimate stoichiommetric acid formed.)
or: Fe +2 + H 2 O + 0.25O 2 = 0.5Fe 2 O 3 + 2H +
FeSO 4 + H 2 O + 0.25O 2 = 0.5Fe 2 O 3 + H 2 SO 4
(g) Check
Left Right Left Right
1Fe 1Fe 0 Charge 0 Charge
1S 1S
5.5O 5.5O
2H 2H OK
This shows that for every one Fe leached then precipitated, 1 mole of H 2 SO 4 is formed = 2 mole H + /mol
CuFeS 2 . Hence 50% of the required acid comes from this step.

2. (a) In the process of question 1, is the purpose of the size reduction steps to expose or to liberate the
chalcopyrite and pyrite minerals (one or the other)?

The purpose is liberation. Fine grinding is used. A concentrate is to be formed, which is mineral separation. This can only be done if the mineral particles are liberated.

(b) A conventional copper process that is very commonly practiced is illustrated in the simplified flowsheet at right.
The acid-leachable copper minerals are typically present at <1% Cu by weight. Is the purpose of the size reduction circuit to expose or to liberate the copper minerals?

The ore is low grade. Grinding is not used, only crushing. The whole ore is leached after stacking onto heaps. All this points to exposure of the desired minerals rather than liberation.

3. A copper ore grading 0.70% copper (average) is crushed for heap leaching. A sample of the crushed ore (989.2 g) was passed through a stack of sieves to determine its size distribution. The data are shown in the table below. The mass retained on each screen is reported. A pan at the bottom of the sieve stack collects any fine material passing through the finest sieve.

\[
\begin{array}{c|c|c}
\text{Tyler number} & \text{Mesh opening }(\mu\text{m}) & \text{Mass retained (g)} \\ \hline
1'' & 25400 & 10.9 \\
0.75'' & 19050 & 61.3 \\
0.5'' & 12700 & 280.9 \\
0.375'' & 9525 & 114.7 \\
0.25'' & 6350 & 128.6 \\
6 & 3360 & 80.1 \\
20 & 841 & 140.5 \\
48 & 297 & 101.9 \\
200 & 74 & 64.3 \\
325 & 44 & 4.0 \\
-325 & {} & 2.0 \\
\end{array}
\]

(a) Calculate the cumulative mass passing each screen size. For instance, for the 1 inch sieve all but 10.9 g passes through. For the 0.75 inch sieve size all but 10.9 + 61.3 g passes, etc.

Data are shown below:

\[
\begin{array}{c|c|c|c|c|c}
\text{Mesh no.} &
\text{Opening }(\mu\text{m}) &
\log(\text{Opening}) &
\text{Mass retained (g)} &
\text{Fraction retained (\%)} &
\text{Cumulative passing (\%)} \\ \hline
1 & 25400 & 4.404834 & 10.9 & 1.1 & 98.9 \\
3/4 & 19050 & 4.279895 & 61.3 & 6.2 & 92.7 \\
1/2 & 12700 & 4.103804 & 280.9 & 28.4 & 64.3 \\
3/8 & 9525 & 3.978865 & 114.7 & 11.6 & 52.7 \\
1/4 & 6350 & 3.802774 & 128.6 & 13.0 & 39.7 \\
6 & 3360 & 3.526339 & 80.1 & 8.1 & 31.6 \\
20 & 841 & 2.924796 & 140.5 & 14.2 & 17.4 \\
48 & 297 & 2.472756 & 101.9 & 10.3 & 7.1 \\
200 & 74 & 1.869232 & 64.3 & 6.5 & 0.6 \\
325 & 44 & 1.643453 & 4.0 & 0.4 & 0.2 \\
-325 & - & - & 2.0 & 0.2 & 0.0 \\
\end{array}
\]

e.g. The total mass is the sum on each screen plus what passed through the finest screen (325 mesh) =
989.2 g. The fraction retained for the no. screen, for instance is 10.9/989.2 = 1.1%.

(b) Plot the cumulative percent passing each screen versus the log 10 of the opening size and determine the P 80 of the crushed ore.

The cumulative mass passing any given screen is everything below that size. For example, the cumulative mass passing in % through the no. 3/8 screen is 13.0 + 8.1 + 14.2 + 10.3 + 6.5 + 0.4 + 0.2 = 52.7%. This is plotted on the figure below:

The P 80 size is indicated on the diagram: log(opening) = 4.2; opening = 10^4.2 = 15,850 μm = 1.59 cm. This means that 80% of the sample is less than 1.59 cm in size.

(c) Assume the ore is mainly a silicate rock with a Bond work index about the same as that of silica (see Introduction to Mineral Processing notes). Calculate the energy requirement for crushing in kWh/ton. For the purposes of the Bond work index equation, assume that the mined ore is very large.

The ore will be reduced in size from large (nominally infinite size) to 15,900 microns. The Bond equation uses size in microns!

\[
W = 10\,W_i\!\left(\frac{1}{d_O^{1/2}} - \frac{1}{d_I^{1/2}}\right)
\]

\[
W = 10 \times 14\,\text{kWh/ton}
\left(\frac{1}{\sqrt{15900}} - \frac{1}{\infty}\right)
\]

\[
W = 10 \times 14
\left(\frac{1}{126.1}\right)
= \textcolor{blue}{1.11\,\text{kWh/ton}}
\]

(d) The as mined ore will be reduced in size to a P 80 of 2.5 cm. Suppose the copper can be 80% leached. Estimate how much power in MW (megawatts) is needed for size reduction if the production rate is to be 50,000 tonnes of copper per year? (1 tonne = 2200 pounds).

Multiplying the feed rate in tons/h times the energy in kWh/ton gives the power needed in kW. The ore
mass flow rate is determined from the copper production rate:

\[
50{,}000\ \text{t Cu prod.}
\times
\frac{1\ \text{t ore}}{0.0070\ \text{t Cu in ore}}
\times
\frac{1\ \text{t Cu in ore}}{0.80\ \text{t Cu prod.}}
\times
\frac{2200\ \text{lb}}{2000\ \text{lb}}
\]

\[
= 9.8214 \times 10^{6}\ \text{tons/year}
= 1121.17\ \text{tons/hour}
\]

The energy requirement:

\[
W = 10 W_i \left( \frac{1}{d_O^{1/2}} - \frac{1}{d_I^{1/2}} \right)
\]

\[
W = 10 \times 14\ \text{kWh/ton}
\left( \frac{1}{\sqrt{25000}} - \frac{1}{\infty} \right)
\]

\[
W = 0.8854\ \text{kWh/ton}
\]

\[
\text{Power}
= 0.8854\ \text{kWh/ton} \times 1121.17\ \text{ton/h}
= 992.7\ \text{kW}
= \textcolor{blue}{0.99\ \text{MW}}
\]

4. A simplified flowsheet for a flotation process on a copper-nickel ore is shown below (p. 7). This process would produce two valuable concentrates, one containing primarily chalcopyrite and the other containing primarily pentlandite, i.e. (Ni,Fe)₉S₈ (the sum of Fe + Ni is 9 moles per 8 moles sulfur). The ore also contains pyrrhotite, i.e. Fe_1-xS (x ≈ 0.05) and other gangue minerals. The ore is crushed to -8 inch at the mine site, then transported to the size reduction/flotation plant. The ore after primary crushing is first screened; -1/2 inch undersize is directed to product. Oversize goes through secondary and tertiary crushing, with -1/2 inch size being the final crushed product. The crushed ore is directed to a two-stage grinding circuit. The ground ore is then sent to flotation for copper and nickel concentrates. There are many possible variations on this theme and actual flowsheets are more complex. Amyl xanthate is commonly used, but for this question assume ethyl xanthate is the collector.

(a) Which parts of the size reduction process are operated in closed circuit fashion, and which in open
circuit manner?

The tertiary crusher and the ball mill are operated in closed circuit mode (product is classified; oversize
is recycled; undersize goes forward).

The gyratory crusher, secondary crusher and rod mill are operated in open circuit mode; there is no
classification and recycle of product.

(b) Is the roughers-scavengers-copper roughers part of the flotation process more like a differential
process or a bulk-selective process?

Two concentrate products are formed: a nickel flotation product and an intermediate copper tailing concentrate. Both are obtained from the same flotation operation after the roughers-scavengers. This is more like a bulk-selective circuit.

(c) Identify a differential circuit within the overall process.

The roughers-scavengers part of the flotation circuit is differential. Most of the ore passes through both stages; two concentrate products are obtained, one from each stage; both flotation products.

(d) Identify one bulk-selective circuit within the overall process.

There are two. Either one will suffice:

Roughers-copper roughers and copper roughers-copper cleaners.

In either case most of the feed (ore in the first case; bulk concentrate in the second) passes through the first stage. Two concentrates are obtained from the second stage, one a flotation concentrate and the other the tailings from that operation.

For the purposes of this discussion, the copper concentrate product the copper roughers is a flotation concentrate product. It then becomes the feed to the cleaners, from which two concentrates are obtained, one a flotation product and the other the tailings.

(e) Which of the following statements are true as concerns how the objectives of good grade and good recovery are achieved in the flotation of the copper concentrate.

(i) Roughers and scavengers employ aggressive flotation conditions as indicated by the extra added xanthate at the scavengers to ensure good recovery of CuFeS₂ , while a subsequent cleaners operation employing mild flotation conditions ensures high grade.

(False - roughers employ moderate conditions for moderate recovery and grade.)

(ii) Sodium cyanide is used to promote aggressive flotation conditions ensuring good recovery.

(False - cyanide is a depressant, not an activator.)

(iii) Moderate flotation conditions are used in the roughers to provide moderate grade and recovery. Scavengers uses aggressive flotation conditions to ensure good recovery.

True

(iv) Ball mills are used in the flotation circuit to effect better liberation of minerals. The ball mill products are recycled through flotation to ensure better recovery and grade.

True

(v) The copper cleaners flotation uses mild flotation conditions with NaCN added as a depressant to ensure that the tailings product from this step has a high grade of CuFeS 2 .

True

(vi) Sodium cyanide is used as a depressant in copper cleaners flotation to ensure that mainly the high-grade CuFeS 2 mineral particles are floated.

True

(vii) Ball mills are used in the flotation circuit to effect better exposure of mineral grains, allowing for better attachment of collector and better recovery.

(False - liberation is the goal, not exposure.)

(f) A pH of 9 is used in rougher flotation to obtain a Ni-Cu concentrate.

(i) How will this affect pyrrhotite in rougher flotation?

From the critical pH curves, a pH of 9 is at the right of the pyrrhotite curve for most xanthate concentrations. Hence pyrrhotite will not float. Right of the curve = does not float. (This can be understood by considering the adsorption equilibria. High pH = increased [OH^-]:

\[ \ce{mineral{\cdot}X^-_{ads} + OH^-_{aq} <=> mineral{\cdot}OH^-_{ads} + X^-_{aq}} \]

(ii) Would the critical pH curve for pentlandite lie to the right or the left of pH 9 and the collector concentration used in this process?

The pentlandite curve should lie to the right of pH 9; a pH below (to the left) of the curve = floats.

(iii) In the course notes no critical pH curve for pentlandite is provided. Do you think the critical pH for pentlandite would lie to the left or the right of the curve for chalcopyrite?

Based on the flowsheet, penatlandite reports to tailings at the pH where chalcopyrite floats. Hence the
pentlandite critical pH should be to the left of the chalcopyrite curve, qualitatively as per the diagram
below:

(g) If the collector concentration was increased, would you expect this shift the critical pH curve for chalcopyrite in the presence of cyanide to lower pH, to higher pH or not at all?

Critical pH curves curve up and to higher pH as collector concentration is increased. This shifts the curve to higher pH. This would occur with or without cyanide present. This makes sense in terms of LeChatelier's principle and the equilibria:

\[ \ce{mineral{\cdot}X^-_{ads} + OH^-_{aq} <=> mineral{\cdot}OH^-_{ads} + X^-_{aq}} \]

or

\[ \ce{mineral{\cdot}X^-_{ads} + CN^-_{aq} <=> mineral{\cdot}OH^-_{ads} + X^-_{aq}} \]