If you add 14.22ml of 2.97m hcl (42.2mmol) to an antacid, then neutralize the excess acid with 5.00ml of 0.1055 m naoh (0.528mmol), how many mmoles of hcl are neutralized by the antacid

Given data:

Volume of HCl = 14.22 ml

Molarity of HCl = 2.97 M

mmoles of HCl = 14.22 * 2.97 = 42.2 mmoles

Volume of NaOH = 5.00 ml

Molarity of NaOH = 0.1055 M

mmoles of NaOH = 5.00 *.1055 = 0.5275 mmoles

Since HCl and NaOH combine in a 1:1 ratio

# moles of NaOH = # moles of excess HCl that is neutralized = 0.5275 moles

Now, the total moles of HCl taken = # mmoles HCl neutralized by antacid + # mmoles of excess HCl

42.2 = mmoles HCl neutralized by antacid + 0.5275

Therefore,

mmoles of HCl neutralized by antacid = 42.2 - 0.5275 = 41.6725 mmoles = 41.7 mmoles

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