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## Q. 3.4

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

## 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