Moles of HCl initially present = M x V ( in
L)
                   Â
= ( 0.0981) x ( 25/1000) = 0.0024525
Moles of HCl titrated with NaOH = NaOH moles
used             Â
( since HCl and NaOH react in 1:1 ratio)
                               Â
= M x V ( NaoH)Â Â = 0.104 x ( 5.83/1000) =
0.0006063
Moles of HCl used in titrating antacid = Initial HCl
moles - HCl moles used in titratiing NaOH
                       Â
= 0.0024525 - 0.0006063Â Â Â Â =
0.0018462
Moles of antacid = Moles of HCl used =
0.0018462
0.204 g sample of ant acid has 0.0018462
moles
Moles of base per gram = ( 0.0018462 /0.204) = 0.00905
moles / g
