Question

Concentration of Standardized HCI Solution = 0.934 M Flask 2 Mass Mg (8) 10.0849 0.0815 Volume HCI (mL) I 10mL 10mL Initial volume NaOH (mL) 12.15ml Final volume NaOH (mL) 21.154L 9 m 21. 15ML 31.5mL 10.35 mL Volume used NaOH (mL) CALCULATIONS **Show work for each trial.** • Moles of Mg used Trial 0.084 Hy loola 10.001g Mg = 0.00 0.003375 mol My 0.0035 molto • Initial moles of HCl (total moles of HCl placed into each Erlenmeyer flask) V: 100L 10*0.934 otma Hal 1000 10 »0.934M 1006 +ICI

Answer 1

Moles of Mg essed in flask 1 - 0.084 molar mass 0.084 24.305 0.00 34 mol. Moles of My used in flask 2 a 0.08! 24.305 = 0.0033 mol

on nitial moles of Hel (Total moles of Hel) antif For flask 1 mass - volume LO clensity 3 1 4 & *2.3333 3 2 1 no of moles – masslused) = 8.3333 0.2286 mol 36.46 molar mass on for flask 2 om oli poi Nos of moles is same as for flask a ie; 0.2286mol. Since two flasks contain 10 ish each.

Moles of NaOH t o Mask 2 tlask a Volume - 21.15 – 12.15 - Aash Jolume 31.5-2115 = 10.35 Mass - Mass! Volume - 10:35 1.515 1.515 I density Volume a density. 1.515 - 5.940591 es elsel mass molar mass - 6.83169 mass molar bass 5.9405 - 6.8316 40 40 - 0. 1485 mol 20.1408 nol

Reaction of Hel and ing. 17 Mg + 2H01 Migel + Ple. Imole of ing react coith amoles of Hein ic, for flasks 0.0034 moh of My react with o.ooo and of Hel Initial mol of HCl = 0.2286 mol. u. remaining moles of Hel reacted with aloob = 0.2286-0.0068 - 080.2218

For flask 2. 0.0033 mol of Mg maid with 0.0066 mol of Hel. .: Remaining mol of Hel marted with Nach - 0.2286 - 0.0066 - 0.222

Moles of Hel that reacted with rig. 2 Hel + Mg - MgCl₂ + H2. No. of moles of the in reacted in flask a - 0 2266 – 0.2218 - 0.0068 No. of moles of the reacted in flask 2 = 0.2286 - 0.222 = 0.0066

UAL CAMERA MI NOTE 75 or . Ratio ano de hon is to flask -1. Ratio. Mokes of Hel reacted in moles of Madison ic, ratio 2:1 for flask-2 0.0068. 0.0034 2 T Ratio - maren 22 0.0063 0.0033 2: ies sho 14

ratio AIDUAL CAMERA REDMI NOTES + Aug . 0.0068 0.0034 0.0063 0.0083 - 1.9545 Ang ratio rounded off is approximately 2.