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  • 11-11-2016
  • Chemistry
contestada

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

Respuesta :

scme1702
scme1702 scme1702
  • 14-11-2016
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
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