Q22.80CP - Chemistry: Molecular Nature Of Matter And Change

Question

The overall cell reaction for aluminium production is 

${\mathbf{\text{2Al}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}(\text{in }\left.{\text{Na}}_{\text{3}}{\text{AlF}}_{\text{6}}\right)\mathbf{\text{ + 3C(graphite)}}\mathbf{\to }\mathbf{\text{4Al(}}\mathit{\ell }{\mathbf{\text{) + 3CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

(a) Assuming 100% efficiency, how many metric tons (t) of ${\mathbf{\text{Al}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}$ are consumed per metric ton of $\mathit{A}\mathit{I}$ produced? (b) Assuming 100% efficiency, how many metric tons of the graphite anode are consumed per metric ton of  produced? (c) Actual conditions in an aluminium plant require 1.89t of ${\mathbf{\text{Al}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}$ and 0.45t of graphite per metric ton of$\mathit{A}\mathit{I}$ . What is the percent yield of  with respect to ${\mathbf{\text{Al}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}$? (d) What is the percent yield of$\mathit{A}\mathit{I}$  with respect to graphite? (e) What volume of  ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(in m}}}^{\mathbf{\text{3}}}\mathbf{\text{)}}$is produced per metric ton of  $\mathit{A}\mathit{I}$at operating conditions of  ${\mathbf{960}}^{\mathbf{°}}\mathit{C}$and exactly1atm?

Step-by-Step Solution

Answer

a) The value is.$\text{1.89 t}$

b) The value is.$\text{0.33 t}$

c) The percentage is .$\text{100\% }$

d) The percentage is $\text{74.2\% }$.

e ) The value is .${\text{2812.7 m}}^{\text{3}}$

Step 1: Define volume

A substance's volume is the amount of space it takes up, whereas its mass is the amount of stuff it contains. The density of a sample is the amount of mass per unit of volume.

Step 2: Explanation

a) We may calculate how many metric tonnes (t) of${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ are consumed. For each metric tonne of $\text{Al}$generated if we assume $\text{100\% }$efficiency then the following can be obtained:

$$\begin{gathered} \text{n}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = }\frac{\text{2}}{\text{4}}\text{n(Al)} \\ \text{n}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = }\frac{\text{1}}{\text{2}}\text{×}\frac{\text{m(Al)}}{\text{A(Al)}} \\ \text{n}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = }\frac{\text{1}}{\text{2}}\text{×}\frac{{\text{1×10}}^{\text{6}}}{\text{26.98gmol-1}} \\ \text{n}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = 18532 mol} \\ \text{m}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = n}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{×M}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right) \\ \text{m}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right){\text{ = 18532 mol×101.957gmol}}^{\text{ - 1}} \\ \text{m}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = 1889492 g} \\ \text{ = 1.89 t} \end{gathered}$$

One metric tonne of $\text{Al}$would consume$\text{ 1.89 }$ metric tonne of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$if the efficiency was $\text{100\% }$.

Therefore, the value is $\text{ 1.89 t}$ .

Step 3: Explanation

b) We may calculate how many metric tonnes (t) of graphite anode are consumed for each metric tonne of produced if the efficiency is assumed to be :$\text{100\% }$

$$\begin{gathered} \text{n(C) = }\frac{\text{3}}{\text{4}}\text{×n(Al)} \\ \text{n(C) = }\frac{\text{3}}{\text{4}}\text{×}\frac{\text{m(Al)}}{\text{A(Al)}} \\ \text{n(C) = }\frac{\text{3}}{\text{4}}\text{×}\frac{{\text{1×10}}^{\text{6}}}{\text{26.98}} \\ \text{n(C) = 27798 mol} \\ \text{m(C) = n(C)×A(C)} \\ {\text{m(C) = 27798mol×12.011gmol}}^{-1} \\ \text{m(C) = 333886 g} \\ \text{ = 0.33 t} \end{gathered}$$

One metric tonne of would require$\text{ 0.33 }$ metric tonnes of C if the efficiency was $\text{100\% }$. (graphite).

Therefore, the value is $\text{ 0.33 t}$

Step 4: Explanation

c) The theoretical mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$required to produce one metric tonne of (at efficiency) is t. As a result, we conclude that 's percent yield to ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$is $100\%$$$\begin{gathered} \text{m}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = 1.89 t} \\ \text{m( graphite ) = 0.45 t } \\ \text{m(Al) = 1 t } \end{gathered}$$.

Therefore, the percentage is $100\%$.

Step 5: Explanation

d) The theoretical amount of graphite necessary to produce one metric tonne of aluminium (assuming $\text{100\% }$efficiency) is$\text{0.33t}$ . The theoretical mass of produced by$0.45$$\text{0.45}$ tonnes of graphite is:

$$\begin{gathered} \text{m}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{ = 1.89 t} \\ \text{m( graphite ) = 0.45} \\ \text{m(Al) = 1 t} \end{gathered}$$

$$\begin{gathered} \text{n(Al) = }\frac{\text{4}}{\text{3}}\text{n( graphite )} \\ \text{n(Al) = }\frac{\text{4}}{\text{3}}\text{×}\frac{\text{m(C)}}{\text{A(C)}} \\ \text{n(Al) = }\frac{\text{4}}{\text{3}}\text{×}\frac{{\text{0.4510}}^{\text{6}}\text{g}}{\text{12.011g/mol}} \\ \text{n(Al) = 49954 mol } \\ \text{m(Al) = n(Al)×A(Al)} \\ \text{m(Al) = 1347764 g} \\ \text{ = 1.35 t } \\ \text{percent yield = }\frac{\text{ experimentally obtained mass of Al }}{\text{ theoretical mass of Al }}\text{×100\% } \\ \text{percent yield = }\frac{\text{1t}}{\text{1.35t}}\text{×100\% } \\ \text{percent yield = 74.2\% } \end{gathered}$$

Therefore, the percentage is$74.2\%$ .

Step 6: Explanation

e) The amount of $C{O}_2$created chemically per metric tonne of Al is:

$$\begin{gathered} \text{n}\left({\text{CO}}_{\text{2}}\right)\text{ = }\frac{\text{3}}{\text{4}}\text{n(Al)} \\ \text{n}\left({\text{CO}}_{\text{2}}\right)\text{ = }\frac{\text{3}}{\text{4}}\text{×}\frac{\text{m(Al)}}{\text{A(Al)}} \\ \text{n}\left({\text{CO}}_{\text{2}}\right)\text{ = }\frac{3}{4}\text{×}\frac{{\displaystyle {\text{1×10}}^{\text{6}}\text{g}}}{{\displaystyle {\text{26.98 gmol}}^{-1}}} \\ \text{n}\left({\text{CO}}_{\text{2}}\right)\text{ = 27798 mol} \\ \text{m}\left({\text{CO}}_{\text{2}}\right)\text{ = n}\left({\text{CO}}_{\text{2}}\right)\text{×M}\left({\text{CO}}_{\text{2}}\right) \\ \text{m}\left({\text{CO}}_{\text{2}}\right){\text{ = 27798mol×44.009 gmol}}^{-1} \\ \text{m}\left({\text{CO}}_{\text{2}}\right)\text{ = 1223362 g} \\ =\text{1.22 t} \end{gathered}$$

We must apply the ideal gas law to compute the volume of$c{o}_2$ :

$$\begin{gathered} T={960}^{°}C \\ \text{ = 1233.15K} \\ \text{p = 1 atm} \\ \text{ = 101325 Pa} \\ \text{pV = nRT} \\ \text{V = }\frac{\text{nRT}}{\text{p}} \\ V=\frac{{\displaystyle {\text{27798mol×8.314JK}}^{-1}{\text{mol}}^{-1}\text{×1233.15K}}}{{\displaystyle \text{101325Pa}}} \\ {\text{V = 2812.7 m}}^{\text{3}} \end{gathered}$$

$2812.7 m^3$Therefore, the value is