A 55.0-mg sample of al (oh) 3 is reacted with 0.200m hcl. how many milliters of the acid are needed to neutralize the al (oh) 3?

Al (OH) ₃ + 3HCl = AlCl₃ + 3H₂O

C=0.200 mmol/mL

M{Al (OH) ₃}=78.0 mg/mmol

m{Al (OH) ₃}=55.0 mg

n{Al (OH) ₃}=m{Al (OH) ₃}/M{Al (OH) ₃}

3n{Al (OH) ₃}=n (HCl) = CV

3m{Al (OH) ₃}/M{Al (OH) ₃}=CV

V=3m{Al (OH) ₃}/[CM{Al (OH) ₃}]

V=3*55.0/[0.200*78.0]=10.6 mL