4.90 Iron metal can be refined (rom the mineral hematite (Fe 2 O 3 ). One way of converting the mineral to iron is to react it with carbon monoxide, as shown below: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2 Because the hematite is obtained from various ores, it is usually not in a pure form. Suppose an iron manufacturer has 2.00 X 10 5 kg of ore available, and the ore is 93% Fe 2 O 3 by mass. (There is no iron in the remaining 7% of the ore.) How many moles of Fe 2 O 3 are present in this ore? How many kg of pure iron could be obtained from this sample of ore? Assume that the process has a 100% yield and that excess CO is available.
4.90 Iron metal can be refined (rom the mineral hematite (Fe 2 O 3 ). One way of converting the mineral to iron is to react it with carbon monoxide, as shown below: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2 Because the hematite is obtained from various ores, it is usually not in a pure form. Suppose an iron manufacturer has 2.00 X 10 5 kg of ore available, and the ore is 93% Fe 2 O 3 by mass. (There is no iron in the remaining 7% of the ore.) How many moles of Fe 2 O 3 are present in this ore? How many kg of pure iron could be obtained from this sample of ore? Assume that the process has a 100% yield and that excess CO is available.
Solution Summary: The author explains how to calculate moles of hematite, mass of pure iron in kg.
4.90 Iron metal can be refined (rom the mineral hematite (Fe2O3). One way of converting the mineral to iron is to react it with carbon monoxide, as shown below:
Fe
2
O
3
+
3 CO
→
2 Fe
+
3 CO
2
Because the hematite is obtained from various ores, it is usually not in a pure form. Suppose an iron manufacturer has 2.00 X 105 kg of ore available, and the ore is 93% Fe2O3 by mass. (There is no iron in the remaining 7% of the ore.) How many moles of Fe2O3are present in this ore? How many kg of pure iron could be obtained from this sample of ore? Assume that the process has a 100% yield and that excess CO is available.
In the phase diagram of steel (two components Fe and C), region A is the gamma austenite solid and region B contains the gamma solid and liquid. Indicate the degrees of freedom that the fields A and B have,
For a condensed binary system in equilibrium at constant pressure, indicate the maximum number of phases that can exist.
Part V. Label ad match the carbons in compounds Jane and Diane
w/ the corresponding peak no.
in the
Spectra (Note: use the given peak no. To label the carbons, other peak
no are intentionally
omitted)
7 4 2
-0.13
-0.12
-0.11
-0.10
-0.08
8
CI
Jane
1
-0.09
5
210
200
190
180
170
160
150
140
130
120
110
100
-8
90
f1 (ppm)
11
8
172.4
172.0
f1 (ppr
HO
CI
NH
Diane
7
3
11
80
80
-80
-R
70
60
60
2
5
-8
50
40
8.
170
160
150
140
130
120
110
100
90
-0
80
70
20
f1 (ppm)
15
30
-20
20
-60
60
-0.07
-0.06
-0.05
-0.04
-0.03
-0.02
-0.01
-0.00
-0.01
10
-0.17
16
15
56
16
-0.16
-0.15
-0.14
-0.13
-0.12
-0.11
-0.10
-0.09
-0.08
-0.07
-0.06
-0.05
-0.04
17.8 17.6 17.4 17.2 17.0
f1 (ppm)
-0.03
-0.02
550
106
40
30
20
20
-0.01
-0.00
F-0.01
10
0
Chapter 4 Solutions
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