Question

Determine how many grams of Al(OH)3 will be required to neutralize 216 mL of 0.367 M HCl according to the reaction:

Answer 1

3HCl + Al(OH)3 > AlCl3 + 3H20

mol = conc × v

= 0.367 × 0.216

= 0.0792 mol HCl

3 mol HCl = 1 mol Al(OH)3

0.0792 mol HCl = x

x = 0.0792/3 × 1

= 0.0264 mol Al(OH)3

Al(OH)3 = 27 + 3(16 +1) = 78 g/mol

mass = mol x molar mass

= 0.0264 × 78

= 2.0592 g

I don't know if it's correct