Question

Concentration of Standardized HCI Solution = 0.39 Flask Mass Mg (g) 0.0845 0.0815 Volume HCI (mL) T 10mL 10mL 12.15 m2 Initial volume NaOH (mL) 21.15ML Final volume NaOH (ml) 21. 15m2 31.5mL Volume used NaOH (mL) 19 mL 10.35 ml CALCULATIONS **Show work for each trial.** . Moles of Mg used • Initial moles of HCl (total moles of HCl placed into each Erlenmeyer flask)

Answer 1

concentration of Hel= 0.39 (M) for Trial-1 For Trim 2 0.084 g of mg = 0.084 mol 0.081 g of Mg = 0.08) = 0.003375 24 of mal mol of me -0.0035 mol ma e volume of All =lome o 10 ml 0.39 m HCl ² loxon 10 and 0.39 an Hel = 10 x 0.39 mol і на = 0.0039 mod ofted = 0.0039 Mol of Hee NaOH needed = (21.15-12.15) - 9 me o NaOH required =(31.5-21.15) = 10.35 me Volume of tee 10 defe Tot is not correct) concentration 0.39 is not correct. a According to dota you have furnished or after reaction of mg, no excess He is a in its (Jo 8. О. оо 50 mc or He is present in both flash For flask 1, 0.007 mod of Hee is I required to reacts with Mg and for flase 2, 0.00675 mol HCl required.

| Mg + 2HCl = mgh + H₂ Imol amol I for I mol of Mg.smed of Helis requered for 0.0035 mol of Mg, no of mal of Hel required = (0.0035x2) = 0.007 mal of Hel (For trial 1) o for 0.003375 mol of Mg, no of mol of tee required = (0.003375x2) = 0.00675 mod of Ace (for trial 2) A Rearaining excess amount of tee = (0.0039-0.007) (Flask 1) =-0.0031 me Remaining excess amount of Hee in flask 2 = Total no of mol of He- no. of inad of Hel reacts with Mg - (0.0039-0.00675) € 0.00285) mol os concentration (0-39M) I think the volume of Hel is some is not correct because lome 0.39M) He means 0.0039 mol of Hel