(a) The balanced reaction is : HI + LiOH ----> LiI +
H2O
(b) Number of moles of LiOH , n = Molarity x volume in L
                                            Â
= 0.250 M x 21.78 mLx 10-3 L/mL
                                            Â
= 0.054 moles
(c) According to the balanced reaction ,
1 mole of LiOH reacts with 1 mole of HI
0.0054 moles of LiOH reacts with 0.0054 moles of HI
(d) The concentration of HI = number of moles of HI / volume of
solution in L
                                    Â
= 0.0054 mol / (34.20 mL x10-3 L/mL)
                                    Â
= 1.592 M
