Question

If $0.5$ moles of oxygen combine with aluminium to form $A{l}_2{O}_3$, then the weight of aluminium metal used in the reaction is:

[atomic weight of $Al=27$ amu]

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

Answer 1

Answer: (B)

$18$

As reaction for the formation of $A{l}_2{O}_3$ is as follows:

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

this shows that 4 mol reacts to 3 mol ( 96 g) of $O_2$

so as 3 mol $O_2$ to 4 mol Al

then 0.5 mol $O_2$ to $\frac{4}{3}\times 0.5=0.66$ mol of Al
as 1 mol Al = 27 g

then 0.66 mol of $Al=0.66\times 27=17.9g\approx 18g$

Hence, the correct option is $B$