Chapter 3
Q. 3.4
GRADED
Metallic iron is most often extracted from hematite ore, which consists of iron(III) oxide mixed with impurities such as silicon dioxide, SiO_{2}.
ⓐ What are the mass percents of iron and oxygen in iron(III) oxide?
ⓑ How many grams of iron can be extracted from one kilogram of Fe_{2}O_{3}?
ⓒ How many metric tons of hematite ore, 66.4% Fe_{2}O_{3}, must be processed to produce one kilogram of iron?
ⓐ
ANALYSIS | |
formula of the iron oxide (Fe_{2}O_{3}) | Information given: |
molar mass (MM) of Fe_{2}O_{3} | Information implied: |
mass % of Fe and O in Fe_{2}O_{3} | Asked for: |
STRATEGY
1. To find the mass percent of Fe follow the plan outlined below. Start with one mole of Fe_{2}O_{3}.
n_{Fe_{2}O_{3}} \underrightarrow{\text{subscript}} n_{Fe} \underrightarrow{\text{MM Fe}} \text{ mass Fe} \underrightarrow{\text{MM Fe_{2}O_{3}}} \% \text{ element }2. Find the mass percent of oxygen by difference. Note that the compound is made up of only two elements, Fe and O.
ⓑ
ANALYSIS | |
mass of Fe_{2}O_{3} (1.000 kg = 1.000 × 10³g) | Information given: |
from (a): mass % of Fe in Fe_{2}O_{3} (69.94% = 69.94 g Fe/100.00 g Fe_{2}O_{3}) | Information implied: |
mass of Fe obtained from 1.000 kg of Fe_{2}O_{3} | Asked for: |
STRATEGY
mass of Fe = (mass Fe_{2}O_{3})(massFe/100 g Fe_{2}O_{3})
ⓒ
ANALYSIS | |
mass of Fe_{2}O_{3} in the ore (66.4% = 66.4 g Fe_{2}O_{3}/100.0 g ore) mass of Fe needed (1.000 kg) |
Information given: |
from (a): mass % of Fe in Fe_{2}O_{3} (69.94% = 69.94 g Fe/100.00 g Fe_{2}O_{3}) factor for converting metric tons to grams. | Information implied: |
mass of hematite needed to produce 1.00 kg of Fe | Asked for: |
STRATEGY
1. Distingush between the mass % of Fe_{2}O_{3} in the ore (66.4%) and the mass % of Fe in Fe_{2}O_{3} (69.94%).
2. Start with 1000 g of Fe and use the mass percents as conversion factors to go from mass of Fe to mass of the ore.
\frac{69.94 g Fe}{100.00 g Fe_{2}O_{3}} \frac{66.4 g Fe_{2}O_{3}}{100.00 g ore}
3. Convert grams to metric tons.
1 metric ton = 1 × 10^{6} g
Step-by-Step
Verified Solution
ⓐ
1 mol Fe_{2}O_{3} × \frac{2 mol Fe}{1 mol Fe_{2}O_{3}} × \frac{55.85 g}{1 mol Fe} = 111.7 g Fe
\frac{111.7 g Fe}{159.7 g Fe_{2}O_{3}} = 69.94% |
mass % Fe |
mass % O = 100% – mass % Fe = 100.00% – 69.94% = 30.06% | mass % O |
ⓑ
1.000 × 10³ g Fe_{2}O_{3} × \frac{69.94 g Fe}{100 g Fe_{2}O_{3}} = 699.4 g Fe | mass Fe |
ⓒ
1000 g Fe × \frac{100 g Fe_{2}O_{3}}{69.94 g Fe} × \frac{100 g ore}{66.4 g Fe_{2}O_{3}} × \frac{1 metric ton}{1 × 10^{6} g} = 2.15 × 10^{-3} metric tons | mass of hematite needed |