Chapter 17, Problem 9PEB

An iron mine has an ore deposit estimated at 3.53 $\times$ 10⁹ metric tons (tonnes). Assays of the deposit indicate it has a hematite content of 56 percent by mass. How many tonnes of iron can potentially be obtained from the deposit? (Assume full recovery.)

To determine

The tonnes of iron can potentially be obtained from the deposit, if it has a hematite content of 56 percent by mass.

Answer to Problem 9PEB

Solution:

$1.38\times {10}^9\text{ tonnes}$

Explanation of Solution

Given data:

An iron mine has an ore deposit of $3.53\times {10}^9\text{ tonnes}$.

Ore grade of the percent is $56\%$.

Formula used:

The mass of hematite in the deposit by multiplying the total mass of the deposit by the ore grade:

$m_{\text{hematite}}=m_{\text{ore}}\left(\text{ore grade}\right)$

The mass of iron by multiplying the mass of hematite by the mass percentage of iron in hematite:

$m_{\text{Fe}}=m_{\text{hematite}}\left(\%{\text{Fe}}_{\text{hematite}}\right)$

Explanation:

Recall the mass of hematite in the deposit by multiplying the total mass of the deposit by the ore grade:

$m_{\text{hematite}}=m_{\text{ore}}\left(\text{ore grade}\right)$

Substitute $3.53\times {10}^9\text{ tonnes}$ for $m_{\text{ore}}$ and $56\%$ for ore grade.

$\begin{array}{c}{m}_{\text{hematite}}=\left(3.53\times {10}^9\text{ tonnes}\right)\left(56\%\right)\\ =\left(3.53\times {10}^9\text{ tonnes}\right)\left(56\right)\left(\frac{1}{100}\right)\\ =1.98\times {10}^9\text{ tonnes}\end{array}$

For Hematite:

The formula: ${\text{Fe}}_2{\text{O}}_3$.

The formula weight of Hematite:

$\begin{array}{llll}Atoms\hfill & Atomicweight\hfill & \hfill & Total\hfill \\ \text{2 of Fe}\hfill & 2\times 28.09\text{ u}\hfill & =\hfill & 111.70\text{ u}\hfill \\ \text{3 of O}\hfill & 3\times 16\text{ u}\hfill & =\hfill & 48.00\text{ u}\hfill \\ \hfill & \text{Formula weight}\hfill & =\hfill & 159.70\text{ u}\hfill \end{array}$

Recall the mass percent of an element in a compound, which can be found from:

$\frac{\left(\begin{array}{l}\text{atomic weight of}\\ \text{ Fe}\end{array}\right)\left(\begin{array}{l}\text{number of atoms }\\ \text{of Fe}\end{array}\right)}{\text{formula}{\text{weight of Fe}}_{\text{3}}{\text{O}}_{\text{4}}}\times 100\%{\text{of Fe}}_{\text{3}}{\text{O}}_{\text{4}}=\%\text{ of Fe}$

Substitute $\left(55.85\text{ u Fe}\right)$ for atomic weight of element, $\left(2\right)$ for number of atoms and $159.70{\text{ u Fe}}_2{\text{O}}_3$ for formula weight of compound.

$\begin{array}{c}\frac{\left(55.85\text{ u Fe}\right)\left(2\right)}{159.70{\text{ u Fe}}_2{\text{O}}_3}\times 100\%{\text{ Fe}}_2{\text{O}}_3=\%\text{ of Fe}\\ \%\text{ of Fe}=\frac{\left(55.85\text{ Fe}\right)\left(2\right)}{159.70}\times 100\%\\ =69.9\text{\% Fe}\end{array}$

Recall the mass of iron by multiplying the mass of hematite by the mass percentage of iron in hematite:

$m_{\text{Fe}}=m_{\text{hematite}}\left(\%{\text{Fe}}_{\text{hematite}}\right)$

Substitute $1.98\times {10}^9\text{ tonnes}$ for $m_{\text{hematite}}$ and $69.9\text{\%}$ for $\%{\text{Fe}}_{\text{hematite}}$.

$\begin{array}{c}{m}_{\text{Fe}}=1.98\times {10}^9\text{ tonnes}\left(69.9\text{\%}\right)\\ =1.98\times {10}^9\text{ tonnes}\left(69.9\right)\left(\frac{1}{100}\right)\\ =1.38\times {10}^9\text{ tonnes}\end{array}$

Conclusion:

The mass of iron is $1.38\times {10}^9\text{ tonnes}$.