Question

Piece +5 (1 pt) 100.00 mL of 1.0 MPb(NO3)2 is added to 150.00 mL of 0.215 M KOH. What is the concentration of KOH? Puzzle #1 (2 pt.) I mix 100.0 mL of 1.00 M Pb(NO3) with 150.0 mL of 0.215 M KOH. What mass of solid product do I expect to recover? Bonus (1 pt): I actually collected 25 g of product. What was the percent yield of my reaction?

Answer 1

100 ml of 1.0M Pb(NO3 is added to 150 ml of 0.215M KOH Vi=isom S, = 0.215M Visi = ₂ 2 W = (100+150 ml S₂ = 150x 0.215 S = ? 100+ 150 = 0.129 M ... The concentration of KOH is 0.129 M. Pb(NO3)₂ (aq) + 2KOH (99) Pb(OH)2 J (s) + 2 KNO3(aq) 100 ml of 1.0M Pb(NO3), with iso ml of 0.215M KOH is mided - 100x mol Pb(NO3)2 with 150X0.215 molkom - - Too = 0.1 mol Pb(NO3), with 0.032 mol Kot is mixed. from the equation it is clean that for 2 mol of KOH when reacts with Imol of Pb(NO3)2, 1 mot of Pb(OH)2 is produced robots sparistot - KOHL-produced tequivalset of P017) 0.032 mol of KOH = 0.032 mol of Pb(OH)2 = 0.016 mol of Pb(OH). Molar mass of Pb(OH)₂ = 241.21 g/mol 0.016 mol of Pb(OH)2 = 241-21*0.0169 = 3.859 g . 3859g of solid mass product you expect to recover, 2.5g of product was collected, % Yield = Actual yield) x 100 (theoritical yield) - 2. 5x100 3.859 = 64.78% • 64.78 % yield of the reaction .