Need help with calculating % yield in Chemistry? Pb(NO3)2(aq) + 2KI(aq) -> 2 KNO
ID: 760454 • Letter: N
Question
Need help with calculating % yield in Chemistry? Pb(NO3)2(aq) + 2KI(aq) -> 2 KNO3(aq) + PbI2(s)? In this equation, determine the amount of precipitate formed when 40ml of potassium iodide reacts with 20ml of lead (II) nitrate. Observations Record weight of lead (II) iodide solid (grams): 9.112g Calculate % yield = (actual yield / predicted yield) x 100% = (actual yield / ((20 ml /1000 ml/liter) x GMW of PbI2)) x 100 % = (actual yield /(.02 x 461.009)) x 100 % = (actual yield / 9.22) x 100% = The potassium iodide reacts with the lead (II) nitrate to yield lead (II) iodide and potassium nitrate. Pb(NO3)2 (aq) + 2KI (aq) --> 2KNO3 (aq) + PbI2 (s) Additional Details lead(II) iodid Pbl2 (In Solution) (0.108000g) (0.000234 moles) Potassium nitrate KNO3 (In Solution) (4.043796g) (0.040000 moles) Water H2O (Liquid) (60.000000g) (3.330558 moles)Explanation / Answer
Need help with calculating % yield in Chemistry? Pb(NO3)2(aq) + 2KI(aq) -> 2 KNO3(aq) + PbI2(s)? In this equation, determine the amount of precipitate formed when 40ml of potassium iodide reacts with 20ml of lead (II) nitrate. Observations Record weight of lead hence we can solve it 4.52 grams of ppt is formed here.