Chapter 26, Problem 26.18P

Interpretation Introduction

Interpretation:

Quantity of iron and slag that will be produced from $10000$ metric tons of ore has to be calculated if the ore contains $50\%{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and $50\%{\text{SiO}}_{\text{2}}$.

Explanation of Solution

Ore of iron is said to contain $50\%{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and $50\%{\text{SiO}}_{\text{2}}$.  Therefore, mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and ${\text{SiO}}_{\text{2}}$ in $10000$ metric tons of ore is calculated as shown below.

    $\begin{array}{c}\text{mass}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{in}\text{10000}\text{metric tons}=10000\text{metric ton}\times \frac{50}{100}\\ =500\text{0}\text{metric}\text{ton}\\ \text{mass}\text{of}{\text{SiO}}_{\text{2}}\text{in}\text{10000}\text{metric tons}=10000\text{metric ton}\times \frac{50}{100}\\ =500\text{0}\text{metric}\text{ton}\end{array}$

Therefore, $\text{5000}\text{metric tons}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is present in $\text{10000}\text{metric tons}$ of ore.  The number of moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ present in $\text{10000}\text{metric tons}$ of ore is calculated as shown below.

    $\begin{array}{c}\text{moles}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{in}\text{10000}\text{metric tons}\text{ore}=\frac{5000\text{metric}\text{ton×}\frac{\text{1}\text{g}}{{\text{10}}^{\text{-6}}\text{metric}\text{ton}}}{159.7\text{g}\cdot {\text{mol}}^{-1}}\\ =3.1309\times {10}^7\text{mol}\end{array}$

Therefore, $\text{5000}\text{metric tons}$ of ${\text{SiO}}_{\text{2}}$ is present in $\text{10000}\text{metric tons}$ of ore.  The number of moles of ${\text{SiO}}_{\text{2}}$ present in $\text{10000}\text{metric tons}$ of ore is calculated as shown below.

    $\begin{array}{c}\text{moles}\text{of}{\text{SiO}}_{\text{2}}\text{in}\text{10000}\text{metric tons}\text{ore}=\frac{5000\text{metric}\text{ton×}\frac{\text{1}\text{g}}{{\text{10}}^{\text{-6}}\text{metric}\text{ton}}}{60.1\text{g}\cdot {\text{mol}}^{-1}}\\ =8.3195\times {10}^7\text{mol}\end{array}$

One formula unit of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ contains two atoms of iron.  This means one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ gives two moles of iron.  Therefore, the moles of iron obtained from $3.1309\times {10}^7\text{mol}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated as shown below.

    $\begin{array}{c}\text{moles}\text{of}\text{Fe}=2\times 3.1309\times {10}^7\text{mol}\\ \text{= 6}\text{.2618}\times {10}^7\text{mol}\end{array}$

Mass of iron obtained from $\text{10000}\text{metric}\text{tons}$ of ore is calculated from the molar mass of iron as shown below.

    $\begin{array}{c}\text{mass}\text{of}\text{iron}\text{obtained}=\text{6}\text{.2618}\times {10}^7\text{mol}\times \text{55}\text{.8}\text{g}\cdot {\text{mol}}^{-1}\\ =\text{349}\text{.40844}\times {10}^7\text{g}\\ =\text{349}\text{.40844}\times {10}^7\text{g×}\frac{{\text{10}}^{\text{-6}}\text{metric}\text{ton}}{\text{1}\text{g}}\\ =3494.1\text{metric}\text{ton}\end{array}$

The mass of iron obtained from $\text{10000}\text{metric}\text{ton}$ of ore is $\text{3494}\text{.1}\text{metric}\text{ton}$.

Slag is formed in the furnace by the reaction between calcium oxide and silica.  This can be represented by chemical equation as shown below.

    ${\text{SiO}}_{\text{2}}+\text{CaO}\to {\text{Ca}}_{\text{2}}{\text{SiO}}_{\text{3}}(slag)$

One formula unit of ${\text{SiO}}_{\text{2}}$ produces one formula unit of ${\text{Ca}}_{\text{2}}{\text{SiO}}_{\text{3}}$.  This means one mole of ${\text{SiO}}_{\text{2}}$ gives one mole of ${\text{Ca}}_{\text{2}}{\text{SiO}}_{\text{3}}$.  Therefore, the moles of ${\text{Ca}}_{\text{2}}{\text{SiO}}_{\text{3}}$  obtained from $8.3195\times {10}^7\text{mol}$ of ${\text{SiO}}_{\text{2}}$ is $8.3195\times {10}^7\text{mol}$.

Mass of slag obtained from $\text{10000}\text{metric}\text{tons}$ of ore is calculated from the molar mass of ${\text{Ca}}_{\text{2}}{\text{SiO}}_{\text{3}}$ as shown below.

    $\begin{array}{c}\text{mass}\text{of}\text{slag}\text{obtained}=8.3195\times {10}^7\text{mol}\times \text{116}\text{.2}\text{g}\cdot {\text{mol}}^{-1}\\ =\text{966}\text{.73}\times {10}^7\text{g}\\ =966.73\times {10}^7\text{g×}\frac{{\text{10}}^{\text{-6}}\text{metric}\text{ton}}{\text{1}\text{g}}\\ =9667.3\text{metric}\text{ton}\end{array}$

The mass of slag obtained from $\text{10000}\text{metric}\text{ton}$ of ore is $9667.3\text{metric}\text{ton}$.

Mass of iron and slag obtained from $10000$ metric tons of ore is $3494.1$ metric ton and $9667.3$ metric ton respectively.