Chapter 18, Problem 18.48QP

(a)

Interpretation Introduction

Interpretation:

Need to calculate the dollars needed for the production of 10 tons of aluminum, 30 tons of sodium and 50 tons of calcium.

Concept introduction:

The reduction reaction for the production of aluminum, sodium and calcium was given below

$\begin{array}{l}{\text{Mg}}^{2+}{\text{+2e}}^{\text{-}}\to \text{Mg}\\ {\text{Al}}^{\text{3+}}{\text{+3e}}^{\text{-}}\to \text{Al}\\ {\text{Na}}^{\text{+}}{\text{+e}}^{\text{-}}\to \text{Na}\\ {\text{Ca}}^{\text{2+}}{\text{+2e}}^{\text{-}}\to \text{Ca}\end{array}$

Even though the above reaction shows the production of different metal but all of them involve the reduction by electrons. The quantity of the metal produced is directly proportional to the quantity of the electrons utilized. So the production cost of the metal depends on the number of moles of electron need.

Answer to Problem 18.48QP

Number of tons of metal produced by one mole of electron was calculated as follows Since the production cost of magnesium was given, the number of tons of magnesium produced by one mole of electrons was calculated first.

For Magnesium

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

$\frac{\text{1}\text{mole}\text{Mg}}{\text{2mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{24}\text{.31g}\text{Mg}}{\text{1}\text{mole}\text{Mg}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=1}{\text{.3298×10}}^{\text{-}}{}^{\text{5}}\text{ton}\text{Mg/mole}{\text{e}}^{\text{-}}$

For Aluminum:

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

$\frac{\text{1}\text{mole}\text{Al}}{\text{3mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{27g}\text{Al}}{\text{1}\text{mole}\text{Al}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=9}{\text{.9206×10}}^{\text{-}}{}^6\text{ton}\text{of}\text{Al/mole}{\text{e}}^{\text{-}}$

Since the production cost of one tone of magnesium was given as 155 dollars it was taken as reference for calculation,

$\begin{array}{l}{\text{Cost of production for Al}}^{\text{3+}}\text{=}\frac{\text{155 dollars}}{\text{1}\text{ton}\text{Mg}}\text{×}\frac{\text{1}{\text{.3298 ×10}}^{\text{-5}}\text{ton}\text{Mg}}{\text{1mole}\text{e-}}\text{×}\frac{\text{1mole}\text{e-}}{\text{9}{\text{.9206 ×10}}^{\text{-6}}\text{ton}\text{Al}}\text{×10}\text{tons}\text{Al}\\ \text{= 2}{\text{.0777×10}}^{\text{3}}\text{dollars}\\ \end{array}$

Explanation of Solution

For Magnesium

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

$\frac{\text{1}\text{mole}\text{Mg}}{\text{2mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{24}\text{.31g}\text{Mg}}{\text{1}\text{mole}\text{Mg}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=1}{\text{.3298×10}}^{\text{-}}{}^{\text{5}}\text{ton}\text{Mg/mole}{\text{e}}^{\text{-}}$

For Aluminum:

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

$\frac{\text{1}\text{mole}\text{Al}}{\text{3mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{27g}\text{Al}}{\text{1}\text{mole}\text{Al}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=9}{\text{.9206×10}}^{\text{-}}{}^6\text{ton}\text{of}\text{Al/mole}{\text{e}}^{\text{-}}$

Since the production cost of one tone of magnesium was given as 155 dollars it was taken as reference for calculation,

$\begin{array}{l}{\text{Cost of production for Al}}^{\text{3+}}\text{=}\frac{\text{155 dollars}}{\text{1}\text{ton}\text{Mg}}\text{×}\frac{\text{1}{\text{.3298 ×10}}^{\text{-5}}\text{ton}\text{Mg}}{\text{1mole}\text{e-}}\text{×}\frac{\text{1mole}\text{e-}}{\text{9}{\text{.9206 ×10}}^{\text{-6}}\text{ton}\text{Al}}\text{×10}\text{tons}\text{Al}\\ \text{= 2}{\text{.0777×10}}^{\text{3}}\text{dollars}\\ \end{array}$

(b)

Interpretation Introduction

Interpretation:

Need to calculate the dollars needed for the production of 10 tons of aluminum, 30 tons of sodium and 50 tons of calcium.

Concept introduction:

The reduction reaction for the production of aluminum, sodium and calcium was given below

$\begin{array}{l}{\text{Mg}}^{2+}{\text{+2e}}^{\text{-}}\to \text{Mg}\\ {\text{Al}}^{\text{3+}}{\text{+3e}}^{\text{-}}\to \text{Al}\\ {\text{Na}}^{\text{+}}{\text{+e}}^{\text{-}}\to \text{Na}\\ {\text{Ca}}^{\text{2+}}{\text{+2e}}^{\text{-}}\to \text{Ca}\end{array}$

Even though the above reaction shows the production of different metal but all of them involve the reduction by electrons. The quantity of the metal produced is directly proportional to the quantity of the electrons utilized. So the production cost of the metal depends on the number of moles of electron need.

Answer to Problem 18.48QP

Number of tons of metal produced by one mole of electron was calculated as follows Since the production cost of magnesium was given, the number of tons of magnesium produced by one mole of electrons was calculated first.

For Magnesium

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

$\frac{\text{1}\text{mole}\text{Mg}}{\text{2mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{24}\text{.31g}\text{Mg}}{\text{1}\text{mole}\text{Mg}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=1}{\text{.3298×10}}^{\text{-}}{}^{\text{5}}\text{ton}\text{Mg/mole}{\text{e}}^{\text{-}}$

For Sodium:

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

$\frac{\text{1}\text{mole}\text{Na}}{\text{1mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{23g}\text{Na}}{\text{1}\text{mole}\text{Na}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=2}{\text{.5353×10}}^{\text{-5}}{}^{}\text{ton}\text{Na/mole}{\text{e}}^{\text{-}}$

$\begin{array}{l}{\text{Cost of production for Na}}^{\text{3+}}\text{=}\frac{\text{155 dollars}}{\text{1}\text{ton}\text{Mg}}\text{×}\frac{\text{1}{\text{.3298 ×10}}^{\text{-5}}\text{ton}\text{Mg}}{\text{1mole}\text{e-}}\text{×}\frac{\text{1mole}\text{e-}}{\text{2}{\text{.5353×10}}^{\text{-6}}\text{ton}\text{Na}}\text{×30tons}\text{Na}\\ \text{=2}{\text{.44×10}}^{\text{3}}\text{dollars}\\ \end{array}$

Explanation of Solution

For Sodium:

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

$\frac{\text{1}\text{mole}\text{Na}}{\text{1mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{23g}\text{Na}}{\text{1}\text{mole}\text{Na}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=2}{\text{.5353×10}}^{\text{-5}}{}^{}\text{ton}\text{Na/mole}{\text{e}}^{\text{-}}$

$\begin{array}{l}{\text{Cost of production for Na}}^{\text{3+}}\text{=}\frac{\text{155 dollars}}{\text{1}\text{ton}\text{Mg}}\text{×}\frac{\text{1}{\text{.3298 ×10}}^{\text{-5}}\text{ton}\text{Mg}}{\text{1mole}\text{e-}}\text{×}\frac{\text{1mole}\text{e-}}{\text{2}{\text{.5353×10}}^{\text{-6}}\text{ton}\text{Na}}\text{×30tons}\text{Na}\\ \text{=2}{\text{.44×10}}^{\text{3}}\text{dollars}\\ \end{array}$

(c)

Interpretation Introduction

Interpretation:

Need to calculate the dollars needed for the production of 10 tons of aluminum, 30 tons of sodium and 50 tons of calcium.

Concept introduction:

The reduction reaction for the production of aluminum, sodium and calcium was given below

$\begin{array}{l}{\text{Mg}}^{2+}{\text{+2e}}^{\text{-}}\to \text{Mg}\\ {\text{Al}}^{\text{3+}}{\text{+3e}}^{\text{-}}\to \text{Al}\\ {\text{Na}}^{\text{+}}{\text{+e}}^{\text{-}}\to \text{Na}\\ {\text{Ca}}^{\text{2+}}{\text{+2e}}^{\text{-}}\to \text{Ca}\end{array}$

Even though the above reaction shows the production of different metal but all of them involve the reduction by electrons. The quantity of the metal produced is directly proportional to the quantity of the electrons utilized. So the production cost of the metal depends on the number of moles of electron need.

Answer to Problem 18.48QP

Number of tons of metal produced by one mole of electron was calculated as follows Since the production cost of magnesium was given, the number of tons of magnesium produced by one mole of electrons was calculated first.

For Magnesium

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

$\frac{\text{1}\text{mole}\text{Mg}}{\text{2mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{24}\text{.31g}\text{Mg}}{\text{1}\text{mole}\text{Mg}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=1}{\text{.3298×10}}^{\text{-}}{}^{\text{5}}\text{ton}\text{Mg/mole}{\text{e}}^{\text{-}}$

For Calcium:

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

$\frac{\text{1}\text{mole}\text{Ca}}{\text{2mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{40g Ca}}{\text{1}\text{mole}\text{Ca}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=2}{\text{.2046×10}}^{\text{-}}{}^{\text{5}}\text{ton}\text{Ca/mole}{\text{e}}^{\text{-}}$

$\begin{array}{l}{\text{Cost of production for Ca}}^{\text{2+}}\text{=}\frac{\text{155 dollars}}{\text{1}\text{ton}\text{Mg}}\text{×}\frac{\text{1}{\text{.3298 ×10}}^{\text{-5}}\text{ton}\text{Mg}}{\text{1mole}\text{e-}}\text{×}\frac{\text{1mole}\text{e-}}{\text{2}{\text{.2046 ×10}}^{\text{-5}}\text{ton}\text{Ca}}\text{×50}\text{tons}\text{Ca}\\ \text{=}\text{4}\text{.675}\times {\text{10}}^{\text{3}}\text{ dollars}\end{array}$

Explanation of Solution

For Calcium:

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

$\frac{\text{1}\text{mole}\text{Ca}}{\text{2mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{40g Ca}}{\text{1}\text{mole}\text{Ca}}\text{×}\frac{\text{1}\text{ton}}{\text{9}{\text{.072×10}}^{\text{5}}\text{g}}\text{=2}{\text{.2046×10}}^{\text{-}}{}^{\text{5}}\text{ton}\text{Ca/mole}{\text{e}}^{\text{-}}$

$\begin{array}{l}{\text{Cost of production for Ca}}^{\text{2+}}\text{=}\frac{\text{155 dollars}}{\text{1}\text{ton}\text{Mg}}\text{×}\frac{\text{1}{\text{.3298 ×10}}^{\text{-5}}\text{ton}\text{Mg}}{\text{1mole}\text{e-}}\text{×}\frac{\text{1mole}\text{e-}}{\text{2}{\text{.2046 ×10}}^{\text{-5}}\text{ton}\text{Ca}}\text{×50}\text{tons}\text{Ca}\\ \text{=}\text{4}\text{.675}\times {\text{10}}^{\text{3}}\text{ dollars}\end{array}$

The quantity of the metal produced is directly proportional to the quantity of the electrons. So the production cost of the metal depends on the number of moles of electron need. Number of tons of metal produce by one mole of electron was calculated first, from this the cost of production of 10 tons of aluminum, 30 tons of sodium and 50 tons of calcium were calculated as $\text{2}{\text{.0777×10}}^{\text{3}}\text{dollars}$, $\text{2}{\text{.44×10}}^{\text{3}}\text{dollars}$ and $\text{4}\text{.675}\times {\text{10}}^{\text{3}}\text{ dollars}$ respectively.