Question

Second run Third run Antacid brand: First run Mass of tablet 1.5939 Volume HCl used 50 ml Molarity of HCI 0.513 M amount of HCl used (total used moles) 25.7 mol Volume NaOH used 75 mL Molarity of NaOH 0.512 M amount of NaOH used (mol) 18.0 mol amount of excess HCl (moles titrated with NaOH) amount of HCl neutralized by antacid (moly to 593g 1.5939 | 50ml 50 mL 0.513 M 0.513 M 25.7 mol 25.7 mol 75 mL 75ml 0.512m 0.512.M 16.7 moll 1607 moll yo 2 amount of CaCOs in tablet (mol) Mass of Caco Mass of CaCO; based on label on package 500mg 2. Difference in mass of CaCO, in tablet (%) Based on packaging information 500 mg 500mg

Answer 1

48 second non 1.593 g 50 ml 0.513 M First nen 1.513 g 50 ml 0.513M asiz möllinoles 75 ml 0.512 Mi 18.0 millimoles 25.7 mmol 75 0:512 16.7 mmol 18.0 millimoles 16.7 mmol Antacid • Mass q lublet • volume of Hclused molarity of Hell · Amount Hclused (total used moe) • volume Naott added molarity of Naolt amount of Naot used (mol) • amount g excess Hel moles titrated with Noote) · amount g Hel nutralised by antacid (mol) . amocent 8 cacąin tablet (mol). • mass of caco mass of calo con label package) | difference in mass of caco (%) since I mole & Hill requires 18 mole of Hil requires 9.O.mmol 7.70 millimeles (7.70x10 mol) 4.5 mmol 3.85 Millimolis 385.4 mg 450 mg 500 mg 90% 77%. & mol of NaOH 18.0 mol of NaOH Ponte tolat = n fick (neutralized by teolie) that fuumalized by NaOH) B = n - n = (25.7 - 18-8 mmol = 7.70 mmolls. ® 1 mol g race, requires a moles y Hel por nuubalization in mol of cacg requires 7.70 molls of Hil = x = 7.70 = 3.85 mmoles & caco o cace = moles x molao mass = 3-85 X100.191 meet = 385.39 mg mass of calls (Mimoles) 20 3