Chapter 17, Problem 10PEB

An ore deposit consisting of chert and hematite occurs in a deposit measuring 2,500 m long by 1.800 m wide by 75 m high. The deposit has an average specific gravity of 4.3. How many tonnes of iron can potentially be obtained from the mine? (Specific gravity can be converted to density by multiplying by the density of water. Assume SG_hematite = 5.3, SG_chert = 2.5. and g/cm³ = tonne/m³.)

To determine

The tonnes of iron can potentially be obtained from the mine, if hematite and chert occurs in a deposit measuring $2500\text{ m}$ long by $\text{1800 m}$ wide by $75\text{ m}$ high.

Answer to Problem 10PEB

Solution:

$6.51\times {10}^8\text{ tonnes}$

Explanation of Solution

Given data:

Specific gravity of hematite is $5.3$.

Specific gravity of deposit is $4.3$.

Specific gravity of chert is $2.5$.

Length of the deposit is $2500\text{ m}$.

Width of the deposit is $\text{1800 m}$.

Height of the deposit is $75\text{ m}$.

Density of deposit is $4.3\frac{\text{tonnes}}{{\text{m}}^3}$.

Formula used:

Write the percent of hematite:

$\%\text{ hematite}=\frac{S{G}_{\text{deposit}}-S{G}_{\text{chert}}}{S{G}_{\text{hematite}}-S{G}_{\text{chert}}}\times 100\%$

Here, $S{G}_{\text{deposit}}$ is the specific gravity of deposit, $S{G}_{\text{hematite}}$ is the specific gravity of hematite and $S{G}_{\text{chert}}$ is the specific gravity is chert.

Write the formula for density $\left(\rho \right)$:

$\rho =\frac{m}{V}$

Here, $V$ is the volume and $m$ is the mass.

Write the formula for volume:

$V=L\times W\times H$

Here, $L$ is the length, $W$ is the width and $H$ is the height.

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 multipling 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 percent of hematite:

$\%\text{ hematite}=\frac{S{G}_{\text{deposit}}-S{G}_{\text{chert}}}{S{G}_{\text{hematite}}-S{G}_{\text{chert}}}\times 100\%$

Substitute $S{G}_{\text{deposit}}$ for $4.3$, $S{G}_{\text{hematite}}$ for $5.3$ and $S{G}_{\text{chert}}$ for $2.5$.

$\begin{array}{c}\%\text{ hematite}=\frac{4.3-2.5}{5.3-2.5}\times 100\%\\ =\frac{1.8}{2.8}\times 100\%\\ =64.3\%\end{array}$

Recall the formula for density $\left(\rho \right)$:

$\rho =\frac{m}{V}$

Rearrange:

$V=\frac{m}{\rho }$

Recall the formula for volume:

$V=L\times W\times H$

Substitute $\frac{m}{\rho }$ for $V$.

$\begin{array}{l}\frac{m}{\rho }=L\times W\times H\\ m=\rho \left(L\times W\times H\right)\end{array}$

Substitute $4.3\frac{\text{tonnes}}{{\text{m}}^3}$ for $\rho$, $2500\text{ m}$ for $L$, $\text{1800 m}$ for $W$ and $75\text{ m}$ for $H$.

$\begin{array}{c}m=4.3\frac{\text{tonnes}}{{\text{m}}^3}\left(2500\text{ m}\times \text{1800 m}\times 75\text{ m}\right)\\ =1.45\times {10}^9\text{ tonnes}\end{array}$

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 $1.45\times {10}^9\text{ tonnes}$ for $m_{\text{ore}}$ and $64.3\%$ for ore grade.

$\begin{array}{c}{m}_{\text{hematite}}=\left(1.45\times {10}^9\text{ tonnes}\right)\left(64.3\%\right)\\ =\left(1.45\times {10}^9\text{ tonnes}\right)\left(64.3\right)\left(\frac{1}{100}\right)\\ =9.32\times {10}^8\text{ tonnes}\end{array}$

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 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 $9.32\times {10}^8\text{ tonnes}$ for $m_{\text{hematite}}$ and $69.9\text{\%}$ for $\%{\text{Fe}}_{\text{hematite}}$.

$\begin{array}{c}{m}_{\text{Fe}}=9.32\times {10}^8\text{ tonnes}\left(69.9\text{\%}\right)\\ =9.32\times {10}^8\text{ tonnes}\left(69.9\right)\left(\frac{1}{100}\right)\\ =6.51\times {10}^8\text{ tonnes}\end{array}$

Conclusion:

The mass of iron is $6.51\times {10}^8\text{ tonnes}$.