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Empirical & Molecular Formula

If 1 1/2 mo...

Question

If $1 \frac{1}{2}$ moles of oxygen combine with $A l$ to form $A l_{2} O_{3}$ , the weight of $A l$ used in the reaction is: $(A l = 27)$

A

27

g

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B

54

g

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C

40.5

g

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D

81

g

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Solution

The correct option is C $54$ g
The molar ratio of Al:O in Al $_{2} O_{3}$ is
$\frac{A l}{O}$ = $\frac{2}{3}$

Therefore Moles of Al = $\frac{2}{3} \times$ Moles of $O_{2}$
= $\frac{2}{3} \times \frac{3}{2} \times 2$
= 2 moles Al

Mass of $A l = 2 \times 27 = 54 g$

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