Question

Consider the following steps involved in the extraction of Aluminium. What is \(Z\)? \[ \text{Bauxite} \xrightarrow[\;523K\;]{\text{Hot conc. NaOH}} X(aq) \xrightarrow[\;CO_2\;]{} Y \xrightarrow[\;1473K\;]{} Z \xrightarrow[\;\text{electrolysis}\;]{} Al \]

• A. \(Al(OH)_3\)
• B. \(Al_2(CO_3)_3\)
• C. \(Al_2O_3\)
• D. \(Al(HCO_3)_3\)

Hint:

In aluminium metallurgy always remember: \[ \text{Bauxite} \rightarrow Al(OH)_3 \rightarrow Al_2O_3 \rightarrow Al \] Electrolysis is carried out on alumina, not directly on bauxite.

Answer 1

The Correct Option is C

Step 1: Understand the flow chart.
We start from bauxite and reach pure aluminium. We must find $Z$, the substance just before electrolysis.

Step 2: First step (forming X).
Bauxite is treated with hot concentrated NaOH at 523 K. The aluminium oxide dissolves to form sodium aluminate. \[ Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4] \] So $X$ is sodium aluminate in solution.

Step 3: Second step (forming Y).
Carbon dioxide is passed through the solution. This brings down aluminium hydroxide as a solid. \[ Na[Al(OH)_4] + CO_2 \rightarrow Al(OH)_3 \downarrow + NaHCO_3 \] So $Y$ is $Al(OH)_3$.

Step 4: Third step (forming Z).
$Al(OH)_3$ is heated strongly at 1473 K. Heating drives off water and leaves pure alumina. \[ 2Al(OH)_3 \xrightarrow{1473K} Al_2O_3 + 3H_2O \] So $Z$ is $Al_2O_3$.

Step 5: Confirm the last step.
Only pure $Al_2O_3$ is melted and electrolysed in the Hall-Heroult process to give aluminium metal. This fits the chart.

Step 6: Identify Z clearly.
The substance fed into electrolysis is alumina, $Al_2O_3$.

Step 7: State the final answer.
Therefore $Z$ is:
\[ \boxed{Al_2O_3} \]