Average Molarity for HCl is .391 Average Molarity for NaOH is .0962

Volume for HCl is:
Trial 1Your 14mL
Trial 2Your 14mL
Trial 3Your 14mL

Volume for NaOH is:
Trial 1: 34.26mL
Trial 2: 33.48mL
Trial 3: 33.84mL

Entry # mass tablet(g) mass antacid(g)Vol HCl(mL)Vol NaOH(mL)
#1: 1.515 0.9010 14.00 34.26
#2: 1.452 0.8370 14.00 33.48
#3: 1.443 0.8280 14.00 33.84

I need help finding the mmoles HCl/mg please.

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl)  is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

the answer is c i believe