Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)?

Question

What volume of 0.470 M HCl is needed to react with 1.95 g of aluminum hydroxide?

Answer 1

Lancenigo di Villorba (TV), Italy

MY ANSWER IS : 0.16 liter.

The reaction involved is

Al(OH)3(s) + 3 HCl(aq) ---> AlCl3(aq) + 3 H2O(aq)

As you know, the Stoichiometric Correspondances are

nAl(OH)3 : n HCl = (-1) : (-3)

In this fashion, that will be

(Al(OH)3's Moles) = 1.95 / 78 = 0.025 mol
(HCl's Moles) = 0.025 * 3 = 0.075 mol
(HCl's Volume) = 0.075 / 0.470 = 0.16 liter

I hope this could be clear.

Answer 2

1. work out the Mr of Al(OH)3
2. 0.47/Mr x 3 x 1000/0.47