Equal weight of Aluminium and Oxygen are allowed to combine with each other to produce ${Al}_{2} O_{3}$ . Identify the correct statement(s).
$(4 Al + 3 O_{2} ⟶ 2 {Al}_{2} O_{3})$ (At wt. of $Al = 27, O = 16$ )

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Aluminium metal is the limiting reagent

The fraction of excess reagent left unreacted is $\frac{1}{9}$ .

The mass of ${Al}_{2} O_{3}$ produced is $\frac{51}{27}$ times the mass of oxygen taken initially.

The mass of aluminium and oxygen is left unreacted and the mass of ${Al}_{2} O_{3}$ produced is double the mass of aluminium taken initially

answer is A, B, C.

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Detailed Solution

$\frac{4 Al}{w} + 3 O_{2} \to 2 {Al}_{2} O_{3}$

$\begin{array}{l} 4 Al + 3 \\ \frac{w}{27} + \frac{w}{32} \\ (L . R) \end{array}$

$- [\frac{W}{32} - \frac{W \times 3}{27 \times 4}] \frac{w}{27} \times \frac{1}{2}$

$= \frac{W}{288}$ $= \frac{W}{54}$ moles

$= 1.9 Wg$

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