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At $25^{0} C, Δ G_{f}^{0}$ of $A l_{2} O_{3} (s) = - 1582 k J m o l^{- 1}, Δ G_{f}^{0}$ of $N a_{2} O (s) = - 377 k J m o l^{- 1} a n d Δ G_{f}^{0}$ of $L i_{2} O (s) = - 544 k J m o l^{- 1} .$ The aluminum oxide can be reduced to Al metal by
(Given $A l_{2} O_{3} (s) + 6 N a (s) \to 3 N a_{2} O (s) + 2 A l (s) a n d A l_{2} O_{3} (s) + 6 L i (s) \to 3 L i_{2} O (s) + 2 A l (s)$ )

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Sodium

Lithium

Both Sodium and Lithium

Can not be predicted

answer is B.

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Detailed Solution

$A l_{2} O_{3} (s) + 6 L i (s) \to 3 L i_{2} O (s) + 2 A l (s) Δ G_{f}^{0} = 3 \times Δ G_{f}^{0} L i_{2} O (s) - Δ G_{f}^{0} A l_{2} O_{3} (s) = 3 \times - 544 - (- 1582) = - 50 k J m o l^{- 1}$

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