Question

If $1.5$ moles of oxygen combines with $Al$ to form $Al_2O_3$, the mass of $Al$ in g [Atomic mass of $Al = 27$] used in the reaction is

• A. $2.7$
• B. $54$
• C. $40.5$
• D. $81$
• E. $27$

Answer 1

Answer: (B)

$\underset{\text { 4moles }}{ 4Al }+\underset{\text { 3moles }}{3 O _{2}} \longrightarrow \underset{2 \text { moles }}{2 Al _{2} O _{3}}$
3 moles of $O _{2}$ combine with 4 moles of $Al$
$\therefore 1.5$ moles of $O _{2}$ will combine with $x mol$ of $Al$
$x =\frac{4}{3} \times 1.5=2\, mol$
Thus mass of $Al$ used in the reaction
$=2 \times 27\, g =54\, g$