Question

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?

Answer 1

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