Concentration of HCl 0.1 M Concentration of NaOH 0.1 M Mass of antacid #1 Run 1)

Question

Concentration of HCl 0.1 M Concentration of NaOH 0.1 M Mass of antacid #1 Run 1) .35 g Run 2) .30 g Run 3) .35 g Origin volume of HCl solution in buret Run 1) 3 mL Run 2) 1 mL Run 3) 2 mL Final volume of HCl solution in buret Run 1) 26.1 mL Run 2) 20 mL Run 3) 26 mL Volume utilized (final - initial) Run 1) 23.7 mL Run 2) 19 mL Run 3) 24 mL Moles of HCl utilized Run 1) mol Run 2) mol Run 3) mol Original volume of NaOH solution in buret Run 1) 0 mol Run 2) 3.8 mol Run 3) 7.2 mol Final volume of NaOH solution in buret Run 1) 3.8 mL Run 2) 7.2 mL Run 3) 11.2 mL

Explanation / Answer

As I understand from the question, the only answer sought is about calculation of number of moles of HCl utilised.

Moles of HCl utilised = Volume of HCl solution used (in L) x molarity of HCl solution

Run 1 :Moles of HCl utilised = (23.7 x 0.1)/1000 = 2.37 x 10-3 moles = 2.37 mmol

Run 2: Moles of HCl utilised = (19.0 x 0.1)/1000 = 1.90 mmol

Run 3: Moles of HCl utilised = (24.0 x 0.1)/1000 = 2.40 mmol

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