Find the volume of 0.100M hydrochloric acid necessary to react completely with 1.51g Al (OH) 3.

Reaction equation:

Al (OH) ₃ + 3HCl → AlCl₃ + 3H₂O

Moles of Al (OH) ₃:

moles = mass/Mr

= 1.51 / (27 + 17 x 3)

= 0.019

Molar ratio Al (OH) ₃ : HCl = 1 : 3

Moles of HCl required = 0.019 x 3

=0.057

concentration = moles/volume

volume = 0.057 / 0.1

= 0.57 dm³

= 570 ml