Question

If $1\frac{1}{2}$ moles of oxygen combine with Al to form $A{l}_2{O}_3$, the weight of Al used in the reaction is: (atomic wt of Al = 27 u)

• A. 27 g
• B. 54 g
• C. 40.5 g
• D. 81 g

Answer 1

The correct option is B 54 g

The balanced chemical equation will be
$3O_2+4Al⟶2A{l}_2{O}_3$
3 moles of $O_2$ combines with 4 moles of $Al$. So $\frac{3}{2}$ moles of $O_2$ will combine with 2 mol of $Al$.
Thus the weight of $Al$ will be $2\times 27=54g$