Chapter 19, Problem 46QP

If the cost of electricity to produce magnesium by the electrolysis of molten magnesium chloride is $155 per ton of metal, what is the cost (in dollars) of the electricity necessary to produce (a) 10.0 tons of aluminum, (b) 30.0 tons of sodium, and (c) 50.0 tons of calcium?

Interpretation Introduction

Interpretation:

The cost of electricity required to form $10.0\text{ tons}$ of aluminum, $30.0\text{ tons}$ of sodium, and $50.0\text{ tons}$ of calcium is to be calculated with given cost of per ton of metal.

Concept introduction:

Electrolysis is thetechnique used to produce a chemical change in the given substance by passing an electric current through it.

Magnesium chloride is the compound of magnesium used in wind erosion. Magnesium is formed by the electrolysis of molten magnesium chloride.

The relationship between grams and $\text{tons}$ can be represented as: $1\text{ ton = 9}\text{.072}\times {\text{10}}^5\text{ g}$

To Convert of $1$ a mole of electron of element into a ton of element, the conversion factor is

$M=\left(\frac{1\text{ mol}\text{element}}{{\text{ mol e}}^{-}}\right)\times \left(\text{molar}\text{mass}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\text{ g}}\right)$.

The cost of electricity to form $\text{tons}$ of element is calculated as: $\text{Cost }=\frac{\text{Cost of electricity per ton of metal}\times \text{moles in per ton Mg}}{\text{Moles in per ton of metal}}\times \text{Amount of metal}$

Answer to Problem 46QP

Solution:

(a)

The cost of electricity to form $10.0\text{ tons}$ of aluminum is $\$2.10\times {10}^3$.

(b)

The cost of electricity to form $30.0\text{ tons}$ of sodium is $\$2.46\times {10}^3$.

(c)

The cost of electricity to form $50.0\text{ tons}$ of calcium is $\$4.70\times {10}^3$.

Explanation of Solution

a) $10.0\text{ tons}$ of aluminum

The ton of the metal formed by $1$ mole of the electron is to be calculated for the reduction reaction.

The reduction reaction is as follows:

${\text{Mg}}^{2+}+2{\text{e}}^{-}\stackrel{}{\to }\text{Mg}$

$1\text{ ton = 9}\text{.072}\times {\text{10}}^5\text{ g}$ and the molar mass of Mg is $24.31\text{ g}$.

Conversion of $1$ a mole of electron of Mg into a ton of Mg is as follows:

$M=\left(\frac{1\text{ mol}\text{element}}{{\text{ mol e}}^{-}}\right)\times \left(\text{molar}\text{mass}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\text{ g}}\right)$

$\begin{array}{l}{M}_{\text{Mg}}=\left(\frac{1\cancel{\text{mol Mg}}}{2{\text{ mol e}}^{-}}\right)\times \left(\frac{24.31\cancel{\text{g}}\text{ Mg}}{1\cancel{\text{mol Mg}}}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\cancel{\text{g}}}\right)\\ M_{\text{Mg}}=1.340\times {10}^{-5}{\text{ ton Mg/mol e}}^{-}\end{array}$

Therefore, one mole of the electrons of Mg produces $1.340\times {10}^{-5}{\text{ ton Mg/mol e}}^{-}$.

The reduction reaction for Al is as follows:

${\text{Al}}^{3+}+3{\text{e}}^{-}\stackrel{}{\to }\text{Al}$

$1\text{ ton = 9}\text{.072}\times {\text{10}}^5\text{ g}$ and the molar mass of Al is $26.98\text{ g}$.

The conversion of $1$ mole of the electrons of Al into a ton of Al is as follows:

$M=\left(\frac{1\text{ mol}\text{element}}{{\text{ mol e}}^{-}}\right)\times \left(\text{molar}\text{mass}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\text{ g}}\right)$

$\begin{array}{l}{M}_{\text{Al}}=\left(\frac{1\cancel{\text{mol Al}}}{{\text{3 mol e}}^{-}}\right)\times \left(\frac{26.98\cancel{\text{g}}\text{ Al}}{1\cancel{\text{mol Al}}}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\cancel{\text{g}}}\right)\\ M_{\text{Al}}=9.913\times {10}^{-6}{\text{ ton Al/mol e}}^{-}\end{array}$

Therefore, per mole of electrons forms $9.913\times {10}^{-6}{\text{ ton Al/mol e}}^{-}$.

The amount given ofAl is $10.0\text{ tons}$.

The cost of electricity to form $10.0\text{ tons}$ of aluminum is calculated as follows: $\text{Cost }=\frac{\text{Cost of electricity per ton of metal}\times \text{moles in per ton Mg}}{\text{Moles in per ton of metal}}\times \text{Amount of metal}$

Substitute the values in the above equation:

$\begin{array}{l}\text{Cost}=\left(\frac{\$155}{1\cancel{\text{ton Mg}}}\right)\times \left(\frac{1.340\times {10}^{-5}\cancel{\text{ton Mg}}}{1\cancel{{\text{mol e}}^{-}}}\right)\times \left(\frac{1\cancel{{\text{mol e}}^{-}}}{9.913\times {10}^{-6}\text{ ton Al}}\right)\times \left(10\text{ ton Al}\right)\\ \text{Cost}=\$2.10\times {10}^3\end{array}$

Thus, the cost of electricity to form $10.0\text{ tons}$ of aluminum is $\$2.10\times {10}^3$.

b) $30.0\text{ tons}$ of sodium

The reduction reaction for Na is as follows:

${\text{Na}}^{+}+{\text{e}}^{-}\stackrel{}{\to }\text{Na}$

$1\text{ ton = 9}\text{.072}\times {\text{10}}^5\text{ g}$ and the molar mass of Al is $22.99\text{ g}$.

The conversion of $1$ mole of the electrons of Na into a ton of Na is as follows:

$M=\left(\frac{1\text{ mol}\text{element}}{{\text{ mol e}}^{-}}\right)\times \left(\text{molar}\text{mass}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\text{ g}}\right)$

$\begin{array}{l}{M}_{\text{Na}}=\left(\frac{1\cancel{\text{mol Na}}}{{\text{1 mol e}}^{-}}\right)\times \left(\frac{22.99\cancel{\text{g}}\text{ Na}}{1\cancel{\text{mol Na}}}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\cancel{\text{g}}}\right)\\ M_{\text{Na}}=2.534\times {10}^{-5}{\text{ ton Na/mol e}}^{-}\end{array}$

Therefore, per mole of electrons forms $2.534\times {10}^{-5}{\text{ ton Na/mol e}}^{-}$.

The amount given ofNa is $30.0\text{ tons}$.

The cost of electricity to form $30.0\text{ tons}$ of sodium is calculated as follows: $\text{Cost }=\frac{\text{Cost of electricity per ton of metal}\times \text{Moles in per ton Mg}}{\text{Moles in per ton of metal}}\times \text{Amount of metal}$

Substitute the values in the above equation:

$\begin{array}{l}\text{Cost}=\left(\frac{\$155}{1\cancel{\text{ton Mg}}}\right)\times \left(\frac{1.340\times {10}^{-5}\cancel{\text{ton Mg}}}{1\cancel{{\text{mol e}}^{-}}}\right)\times \left(\frac{1\cancel{{\text{mol e}}^{-}}}{2.534\times {10}^{-5}\text{ ton Na}}\right)\times \left(30\text{ tons Na}\right)\\ \text{Cost}=\$2.46\times {10}^3\end{array}$

Thus, the cost of electricity to form $30.0\text{ tons}$ of sodium is $\$2.46\times {10}^3$.

c) $50.0\text{ tons}$ of calcium

The reduction reaction for Ca is as follows:

${\text{Ca}}^{2+}+2{\text{e}}^{-}\stackrel{}{\to }\text{Ca}$

$1\text{ ton = 9}\text{.072}\times {\text{10}}^5\text{ g}$ and the molar mass of Ca is $40.08\text{ g}$.

The conversion of $1$ mole of the electrons of Ca into a ton of Ca is as follows:

$M=\left(\frac{1\text{ mol}\text{element}}{{\text{ mol e}}^{-}}\right)\times \left(\text{molar}\text{mass}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\text{ g}}\right)$

$\begin{array}{l}{M}_{\text{Ca}}=\left(\frac{1\cancel{\text{mol Ca}}}{{\text{1 mol e}}^{-}}\right)\times \left(\frac{40.08\cancel{\text{g}}\text{ Ca}}{1\cancel{\text{mol Ca}}}\right)\times \left(\frac{1\text{ ton}}{9.072\times {10}^5\cancel{\text{g}}}\right)\\ M_{\text{Ca}}=2.209\times {10}^{-5}{\text{ ton Ca/mol e}}^{-}\end{array}$

Therefore, per mole of electrons forms $2.209\times {10}^{-5}{\text{ ton Ca/mol e}}^{-}$.

The amount given ofCa is $50.0\text{ tons}$.

The cost of electricity to form $50.0\text{ tons}$ of calcium is calculated as follows: $\text{Cost }=\frac{\text{Cost of electricity per ton of metal}\times \text{moles in per ton Mg}}{\text{Moles in per ton of metal}}\times \text{Amount of metal}$

Substitute the values in the above equation:

$\begin{array}{l}\text{Cost}=\left(\frac{\$155}{1\cancel{\text{ton Mg}}}\right)\times \left(\frac{1.340\times {10}^{-5}\cancel{\text{ton Mg}}}{1\cancel{{\text{mol e}}^{-}}}\right)\times \left(\frac{1\cancel{{\text{mol e}}^{-}}}{2.209\times {10}^{-5}\text{ ton Ca}}\right)\times \left(50\text{ tons Ca}\right)\\ \text{Cost}=\$4.70\times {10}^3\end{array}$

Thus, the cost of electricity to form $50.0\text{ tons}$ of calcium is $\$4.70\times {10}^3$.