scastrowilson3829 scastrowilson3829
  • 30-11-2021
  • Chemistry
contestada

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

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lilmhz
lilmhz lilmhz
  • 10-12-2021

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

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