Chemistry,  moles neutralization enthalapy heat capacity?

1.
(a)
Consider the reaction of the excess HCl with NaOH:
HCl + NaOH → NaCl + H₂O
Mole ratio HCl : NaOH = 1 : 1

Moles of NaOH reacted = (0.1 mol/L) × (17.5/1000 L) = 0.00175 mol
Moles of excess HCl = 0.00175 mol

Consider the reaction of the HCl with antacid:
[4MgCO₃•Mg(OH)₂•H₂O] + 10 HCl → 5 MgCl₂ + 4CO₂ + 7H₂O
Mole ratio MgCO₃ : Mg(OH)₂ : HCl = 4 : 1 : 10

Total moles of HCl added = (0.15 mol/L) × (50/1000 L) = 0.0075 mol
Moles of HCl reacted with antacid = (0.0075 - 0.00175) mol = 0.00575 mol

In the antacid:
Moles of MgCO₃ = (0.00575 mol) × (4/10) = 0.0023 mol
Moles of Mg(OH)₂ = (0.00575 mol) × (1/10) = 0.000574 mol

====
(b)
(i)
Moles of NaOH added = (2.5 mol L⁻¹) × (70/1000 L) = 0.175 mol
Moles of HCl added = (3.2 mol L⁻¹) × (55/1000 L) = 0.176 mol

1 mol NaOH reacts with 1 mol HCl to form 1 mol of H₂O.
HCl is slightly excess, and NaOH is the limiting reactant (limiting reagent).
Moles of H₂O formed = 0.175 mol

Mass of NaOH solution = (70 mL) × (1.04 g mL⁻¹) = 72.8 g
Mass of HNO₃ solution = (55 mL) × (1.51 g mL⁻¹) = 83.05 g

q = m c ΔT
(68 × 1000 J/mol) × (0.175 mol) = [(72.8 + 83.05) g] × (4.18 J g⁻¹ °C) × (T - 32)°C
68000 × 0.175 = 155.85 × 4.18 × (T - 32)
T - 32 = 68000 × 0.175 / (155.85 × 4.18)
T = 50
Highest temperature = 50°C

(ii)
There is no change in the molar enthalpy of neutralization.
By definition, molar enthalpy of neutralization
= (Heat evolved)/(Moles of water formed)
When higher concentration of NaOH and HNO₃ solutions were used, the heat evolved and the number of moles of water formed would both increase to the same extent, and thus the molar enthalpy of neutralization would remain constant.

hmmm... phenolphthalein has a pKa = 9.4.  so we could do this the easy way or the complicated way.  Have you covered pH of buffered solutions yet?