Question

How many moles ofOH- are there in 4 moles ofAl(OH)3

Answer 1

Solution.

$AL(OH)3 = Al^{3+} + 3OH^{-}$

$n(OH^{-}) = 3 \times n(Al(OH)3)$

$n(OH^-) = 3 \times 4 = 12 \ mol$

Answer:

$n(OH^-) = 3 \times 4 = 12 \ mol$