If 4.5 gm Aluminium completely reacts with 4gm of oxygen. Then what will be empirical formula of aluminium oxide.
A. `Al_(2)O`
B. `Al_(2)O_(3)`
C. `AlO_(2)`
D. `AIO`

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1 Answers

`Al+O_(2)rarrAl_(x)O_(y)`
`n_(Al)=(4.5)/(27)=(1)/(6) n_(o_(2))=(4)/(52)=(1)/(8) (n_(Al))/(n_(o))=(1)/(6)xx(4)/(1)=(2)/(3)`
`n_(o)=2xxn_(o_(2))=(1)/(4) "Formula"=Al_(2)O_(3)`

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