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# If half mole of oxygen combine with Al to form $Al_2O_3$,the weight of Al used in the reaction is

$(a)\;27g\qquad(b)\;40.5g\qquad(c)\;54g\qquad(d)\;18g$

$4Al+3O_2\rightarrow 2Al_2O_3$
Ratio of weight of $Al$ and $O_2$ is 9 : 4
$\therefore \large\frac{9}{4}=\frac{x}{8}$$=18g$
Hence (d) is the correct answer.