Find the volume of 0.110 m hydrochloric acid necessary to react completely with 1.53 g al(oh)3.

Solution:

Molarity is defined as moles per volume.

So,

Since,We have 1.53g of the Aluminium complex which makes (moles =mass/mr) (1.53/78) = 0.019618 moles of aluminium complex The fundamental aspect to know is that one HCl will react with one of the OH molecules on the complex and so for each mole of Al(OH)3 you need 3HCl...  

The full reaction is:- Al(OH)3 + 3HCl - - > AlCl3 + 3H20. So 0.019618 X 3 = 0.0588 moles of HCl needed, and if we then plug that back into the first equation rearranged for volume we arrive at the answer which is 0.0588 moles / 0.110 = 0.534 dm3 or 534ml.