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22 October, 21:57

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

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  1. 22 October, 23:24
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    Given dа ta:

    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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