Question

with the information provided please answerr questions 8-14

Table 1 - Generic Antacid Hydrochloric Acid (Acid Flask Preparation) 1 Initial reading of buret filled with 0.10 M Ha 2 Final reading of buret filled with 0.10 M HC 3 Volume of 0.10 M HCl delivered to acid flask Data Set 1 Data Set 2 22 12,8749.35 ML 24.37 m2 24.48mL Antacid 4_Actual mass of antacid added to add flask 0.IZI 0 123 5 5 6 7 Titration with Sodium Hydroxide Initial reading of buret filled with 0.10 M NaOH Final reading of buret filled with 0.10 M NaOH Volume of 0.10 M NaOH used in titration - Number of moles of Hal initially in the acid flask Show work 0,39 M S AM 15.30mL ML S INO MLL 5.41 mL Number of moles of NaOH used in the titration Show work 10 Number of moles of HCl neutralized by 0.10 M NaOH during the titration. (The antacid has already neutralized some HCl. The titration neutralized the rest.) Show work: Titration III - Report Sheet (age 2]

Answer 1

Moles = molarity x volume in litres. IML = 0.001L Data seti Data set 2. S.NO 8. Moles of Hce initially in the acid flasf O moles - 0.10M X 0.024371. - 2 moles = 0.10 M4 0.024482= 0.002437 moles 0.002448 moles. 0.001530 moles 0.001541 moles. Number of moles of NaOH used in the titration; 0 0.10M X 0.015304 - ② 0.10 mx 0.01541L = He & NaOH neutralisation equ. Hel + NaOH = NaCl + H₂O I mol NaOH neutralises Imol of 1 10. 0.001530 moles 0.001541 moles. He 0.002437 -0.001530 0.002448 -0.001541 So, moles of NaOH used = moles of Hel neutralised by it. moles of HCl consumed by CaCO3 in antacid sample; (total initial HCl moles - moles neutralised by NaOH) Chemical equi. I CaCO₃ + 2HCl = CaCl₂ + H₂O +(0, Thus, = 0.000907 moles = 0.000907 moles 12. 0.000907 moles Hul 0.0004535 moles. 2 moles of CaCO₃ = 1 mol. Hel = 0.0004535 moles CaCO₃. 13. 0.04539g 14. Crams of Cacoz in sample; 0.0004535 miles x 100.089 molesxmolar mass of CaCO3 mot = 0.04539g 7. CaCO3 in sample; 0.045399x100 mass of caroz as calculated mass of antacid added J = 37.51 1. CaCO3 Kindly rate the answer!!!! 0.04539gx100 mass of Calog as 13 x100 0.1219 0. 1239 - 36.9%. CaCO3