Question

The half mole of oxygen combine with $Al$ to form $A{l}_2{O}_3$, the weight of $Al$ used in the reaction is:

• A. $27g$
• B. $40.5g$
• C. $54g$
• D. $18g$

Answer 1

The correct option is D $18g$

$4Al+3O_2\to 2A{l}_2{O}_3$

$3$ mole $O_2$ combines with$\to 4\times 27g=108gof$ $Al$

So, $0.5$ mole combines with $=\frac{108\times 0.5}{3}=18g$