Question

If $1.5$ moles of oxygen combine with $Al$ (atomic mass $=27$ g) to form $A{l}_2{O}_3$, the weight (in g) of $Al$ used in the reaction is:

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

Answer 1

Answer: (B)

$54$

$2Al+1.5O_2\to A{l}_2{O}_3$

Thus, $2$ mol of $Al$ combine with $1.5$ mol of oxygen to form $1$ mol of $A{l}_2{O}_3$.

$2$ mol of $Al$ corresponds to $2\times 27=54$ g.

Thus, the weight of $Al$ used in the reaction is $54$ g.

Hence, the correct option is $B$