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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Ответ:
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
Ответ:
Answer is 1.41 x 10 with the exponent of 24, atoms
Explanation : Look at the picture i attached.