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)

Answer

B

Explanation

`underset(4 xx 27 = 54 g)(4Al) + underset(3 mol)(3O_(2)) to 2Al_(2)O_(3)`
Amount of Al that combines with 3 moles of `O_(2) = 54 g`
`therefore` Amount of Al that combine with 1.5 moles of `O_(2) = 27 g`