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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