If `1 1/2` moles of oxygen combine with Al to form `Al_(2)O_(3)` the weight of Al used in the reaction is (Al=27)

Options

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

Answer

B

Explanation

`{:(3O_(2)+4Alrarr2Al_(2)O_(3)O),(3rarr4),(1rarr(4)/(3)xx(3)/(2)),((3)/(2)rarr=2molxx27=54 g):}`