Question

The mass of carbon anode consumed (giving there only carbon dioxide) in the production of ${\rm{270}}\,{\rm{Kg}}$ of aluminium metal from bauxite by the Hall process is:
$\left[ {{\rm{Atomic}}\,{\rm{mass:}}\,{\rm{Al}}\,{\rm{ = }}\,{\rm{27}}} \right]$
(A) $90\,{\rm{Kg}}$
(B) $540\,{\rm{Kg}}$
(C) $180\,{\rm{Kg}}$
(D) ${\rm{270}}\,{\rm{Kg}}$

Answer 1

: As we know that bauxite is the ore of aluminium metal and represented in the aluminium oxide. In the Hall process, this aluminium oxide gives pure aluminium metal by reacting with carbon. This process undergoes electrolysis. So, with the help of reaction, we can calculate mass of carbon anode by calculating $1\,{\rm{Kg}}$ of aluminium metal gives how much kilogram of carbon.

Complete answer:
Now, come to our answer, the reaction occurs in Hall process is-
${\rm{2}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\,{\rm{ + }}\,{\rm{3}}\,{\rm{C}}\,\, \to \,{\rm{4}}\,{\rm{Al}}\,{\rm{ + }}\,{\rm{3}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}$
In this reaction, as we can see that four moles of aluminium is produced by three mole of carbon or we can represent it in kg also by unit conversion as-

{\rm{2}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\,\,\,\,\,\,{\rm{ + }}\,\,\,\,\,\,{\rm{3}}\,{\rm{C}}\,\,\,\, \to \,\,\,\,\,\,\,{\rm{4}}\,{\rm{Al}}\,{\rm{ + }}\,{\rm{3}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{3}}\,{\rm{ \times }}\,{\rm{12}}\,{\rm{g}}\,\,\,\,\,\,\,\,\,{\rm{4 \times }}\,{\rm{27}}\,{\rm{g}}\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\rm{.036}}\,{\rm{Kg}}\,\,\,\,\,\,\,{\rm{0}}{\rm{.108}}\,{\rm{Kg}} \end{array}$$ So, $$0.108\,{\rm{Kg}}$$ of aluminium metal produce $${\rm{0}}{\rm{.036}}\,{\rm{Kg}}$$ of carbon at anode So, $$1\,{\rm{Kg}}$$ of aluminium metal will produce $$\begin{array}{l} {\rm{ = }}\,\dfrac{{{\rm{0}}{\rm{.036}}\,{\rm{Kg}}}}{{{\rm{0}}{\rm{.108}}\,{\rm{Kg}}}}\,\\\ {\rm{ = }}\,{\rm{0}}{\rm{.33}}\,{\rm{kg}} \end{array}$$of carbon Therefore, $${\rm{270}}\,{\rm{Kg}}$$ of aluminium metal will produce $$\begin{array}{l} {\rm{ = }}\,\dfrac{{{\rm{0}}{\rm{.036}}\,{\rm{Kg}}}}{{{\rm{0}}{\rm{.108}}\,{\rm{Kg}}}}\,{\rm{ \times }}\,{\rm{270}}\,{\rm{Kg}}\\\ {\rm{ = }}\,{\rm{90}}\,{\rm{Kg}} \end{array}$$ of carbon. **Hence the correct option is (A).** **Note:** : The given method of answer is one of the easiest methods to calculate this type of questions, only we have to know about the reaction and unit conversion. It is applicable to calculate mass of reactants, mass of products, moles of reactants and products.