Question

please answer 9-14.

Table 1 - Generic Antacid 1 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 HCI 3 Volume of 0.10 M HCl delivered to acid flask Data Set 1 Data Set 2 0.50 ml 24. 1872935 ml 24.37m2 24.48mL 4 Antacid Actual mass of antacid added to acid flask 0. 12 0 .123 5 6 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 0.31 m 16.IML 5. ML 31.10 ML 15.41 mL 115.30mL Number of moles of HCl Initially in the acid flask Show work Number of moles of NaOH used in the titration 10 Number of moles of HCI neutralized by 0.10 M NaOH during the titration. (The antacid has already neutralized some HCI. The titration neutralized the rest.) Show work: Titration - Report Sheet(page 2]

Answer 1

Date: Expt. No. Experiment: molanty of Hal = 0.10m Set-1 Thuret Initial reading = 50ml final reading = 24.87 ml volume was used = 24.37 ml during tistration No. moles of Hal = molarity & volume = 0.10m x 24.31ml. at loomL- 1L 0.10m x 24.37 L 2.4x103 moles =No. of mole used - acid flaske weight of auto acid & ursed 0.121 g lo ой - о јом buret Initial reeding - 0 39 mL final velay = 15.69 ml rano Volume eyed 15.30 mL No. moles of NaOH = 15.30 ml x 0.10 M = 1.53 xom hole

set-2 Pata 93 No. moles of Hal = 24.48mL xo.1om - 2.448 pinole Vaot = 15.4ink X D.Om rono = 1.54 1x mole. Data set- No. of moles of Hol = 1.54x10* Data Set a Hoof mote to to of motos of the +594 mote as the 20 1 mole of Nooh will neutralized the 1 mole of Hel. NaOH is added offer the antacid . No. of moles of Hal = 1.53 1103 Data set - 2 Get 1 2.400 x 103 mole Hed 1.53 x 103 mole hal 2.448 x 10litet ole tea 1.541 x 103 molo Her 0.907 x 10-3 mole Hah 0.870 x 10 mole Hal 2 Hal + CaCO3 Calle + H₂CO3 Ben 2 mole of had react to give Imole a coz Set 1 0.870 x 103 mole Mal x 2 = 0.43syanole of cu coz Set 2 = 0.9078103 mole Hell x 2 = 0.45 35x10 more of card, So.

Expt. No. Experiment: | Date: Date: Dronret 3. molecular weight of CaCO₃ = 100 g/mol So. Dat a set 1 Other brands from jn raper were grams of Caloz = mole of a CO₂ x molecular weight = 0.1435x103 mole x 100 g/mol 0.435 x 10-7 0.043 g Data set 2 grams of Cacos = 0.4535x103 x lov = 0.0453 g. percentage of calcium CaCO₃ = weight of CaCO3 ing xin total weight of Sample! Data set - 1 a % of CaCO₃ = 0.0437 x lov 0.9219 Data set 2 of = 35.53 % of Caco, = 0.04539 x 100 0:1239 = 36.82%