Question

How was 23 chosen as the actual value for the percentage yield?

2. For the reaction: Cu2O(s) + C(s) 2 Cu(s) + CO(g) When 11.5 mol of Care allowed to react with 145 mol of Cu20 10.0 mol of Cu are obtained. Determine (a) the limiting reactant and the theoretical yield of copper, and then (b) the percent yield. Show the equation for percent yield first and then include the values in this equation and solve. Xtremting Reactant - amal Cu 23.0 mal Co 17.5 mal c Imal C L U ly didyau choose ©) 14.5 nel Cuso Tual Cuz 07 % field = actual x100 => 10-x100=43.5% 23nd Co= Theorehead and Co-actual 79 Chemistry 101 Workbook Part II: Lecture Worksheets

Answer 1

way ---1
Cu₂ 063) + ((6) 2 Ca(s) + Colg) Stempo find Limiling keagent. a) mole here Coffre of CuO = males ont Cofficient - 1405mol - 1405 b) males per Coffo of C= 11.5 mol = 1105 : 11.5 <1405 80. C acts as Limiting Reagend steho applied Done Rule. moles of Cu moles of C Cofficient officient. moles of Ca = 1105 melés ofla = 23 mol (thewritical gesla). Step 1 % feild = actual geild x 100 Theoritical geld % geild = 10 x 100 = 43.48 % dun, Thank you, May it help you.
or way-2
DATEITIT Cuz o co + Imole con Imole 2urs + cocg). mote Imole I mele Cuzo = I mele ( o • 14.5 mole of Cyo required = 14.5 mole 'C' so Carbon is limiting Reagent Theoretical yield we know that limiting Reagent I decide the Amount of Product According to Reaction . 1 mole Carbon Produce = amole cu Now 11.5 mole Carbon Prochuce = 2x115 Theoretical field = 23 mo (6) actual yield A Dercent yield = x100 yield theoretical 10 X 100 =43 23 PAGE
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