Question

Section: Due Date: 10/31/19 INSTRUCTION: Give your answers in the space provided. You MUST show your calculations to get credit. ++++++ ++++++++++++++++++++++++++++++ +++++++++++++++++ 1. Calculate the mass (in g) of AgCl that can be formed from the reaction of 85.0 mL of 0.110 M MgCl, solution with 65.0 mL of a 0.150 M AgC,H,O, solution. (2 Pts.) MgCl2(aq) + AgC,H,O,(aq) AgCl(s) + Mg(C,H,02), (aq) 2. Calculate the molarity of a solution prepared by dissolving 2.500 g of Na2CO, in water to make 500 mL of solution. (2 pts.)

Answer 1

solution ① Given neaction. mg clacaq) + Ag C₂ H₂O2 (aq) → AgCl, so + mg (C2H2O2), la mmol of machz = 85.0 mLyo.110M 29.35 mmol m mol. of Ag C₂H₂O2 = 650ml x 0.150 19 29.75 mmol mmol. of Mach is less than the mmol of Ag C2H₂02 So mgch is limiting reagent so. mmol of All produced a 9.35 mmol since. I'm mol 2103 mol No. of mol of Agee = 9.35 mol 103 20.00 935 mol As you know that mass No. of moles nodar mass therefore mass of Agll a 2 = No-fmoles of Ageix molarmetsch Arle 0.00935% 143.32 .. 1.34 g (Asis)

solution ③ We know the molarity definition. - Molarity z No. of modes solute volume of solution (in Lj And No. of moles = mars Molermass given mass of Naz CO₂ 22.500g Molar mass of Na2CO3 = 105.98 g/mol, . No. of moles z 2.500g Na₂CO3 105.98 g/mol - 2.500g -0.0236 mol volume of solution a soont 2 500 m (:1m1=163 ) -3 IML =10 2 500 ml דערטן 20.5 L Therefore, Molaroty z 0.0235 mol 05 (L) = 0.047 mol 20.047 m