Answered: An antacid tablet was ground, and 0.510…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Transcribed Image Text:o ono ol bibs pholdsbyr Antacid Cost per brand and number of tablets: Tums (CaCO3 500mg) Price: $8.37/pack (1 pack has 150 tablets) Alka Seltzer (NaHCO3 1936 mg) Price: $8.75/pack (1 pack has 36 tablets) Alcalak (CaCO3 420 mg) Price: $23.06/pack (1 pack has 500 tablets) Sample calculation: An antacid tablet was ground, and 0.510 g of the powder was dissolved in 30.00 mL of 0.500 M HCl. The excess acid was back-titrated with exactly 9.25 mL of 1.10 M NaOH. The average weight of a tablet is 1.105 g. The tablet came from a bottle of 200 tablets that cost $4.99. anoitsipl a) Calculate the moles of HCl neutralized per tablet. mon b) Calculate the mass effectiveness of the antacid. c) Calculate the cost effectiveness of the antacid. a) Calculate the moles of HCl neutralized per tablet. Calculate the number of moles of HCl added to the sample Moles HCl = (0.03000L) (0.500M) = 0.0150 mol Calculate the number of moles of NaOH required to back-titrate the excess HCI Moles NaOH = (0.00925L) (1.10M) = 0.010175 mol Calculate the number of moles of HCl neutralized per sample Moles HCl neutralized = moles of HCl - moles of NaOH = 0.0150 mol-0.010175 mol = 0.004825 mol Calculate the number of moles of HCl neutralized per tablet: 0.004825 mol x 1.105 g/0.510 g = 0.01045 mol b) Calculate the mass effectiveness of the antacid. moles of HCl neutralized per tablet = 0.01045 mol mass of antacid = 1.105 g E = moles of HCl neutralized per tablet / mass of antacid = 9.5 x 10-3 mol/g c) Calculate the cost effectiveness of the antacid. moles of HCl neutralized per tablet = 0.01045 mol price per tablet, ¢ = ($4.99/bottle) (100¢/1$)(1 bottle/200 tablets) = 2.495 c C = moles of HCl neutralized per tablet / price per tablet = 4.2 x 10-3 mol/¢

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