`0.078 g Al(OH)_(3)`. is dehydrated to `Al_(2)O_(3)`. The `Al_(2)O_(3)` so obtained reacted with `6` milli-equivalent of `HCl`. The equivalent of `AlCl_(3)` produced during the reaction are:

Options

A. `10^(-3)`
B. `3xx10^(-3)`
C. `4xx10^(-3)`
D. `(10^(-3))/(2)`

Answer

B

Explanation

`Al(OH)_(3)rarrAl_(2)O_(3)overset(6.Meq "of" HCl)rarrAlCl_(3)`
Meq. of `Al(OH)_(3)=(0.078)/(78//3)xx10^(3)=3`
Meq. Of `AlCl_(3)`formed`=3 (HCl` in excess)
Eq. of `AlCl_(3)` formed `= 3xx10^(-3)`