A reaction of 345.0 g of C 2 H 6 (g) with600.0 g of O 2 gas produces CO 2 (g) an

Question

A reaction of 345.0 g of C2H6 (g) with600.0 g of O2 gas produces CO2 (g) andH2O (g). a. How much CO2 is produced if all theC2H6 is used? b.What is the limiting reagent? c.What is the theoretical yield? d.If the actual yield is 386.0 g of CO2 , what isthe percent yield? A reaction of 345.0 g of C2H6 (g) with600.0 g of O2 gas produces CO2 (g) andH2O (g). a. How much CO2 is produced if all theC2H6 is used? b.What is the limiting reagent? c.What is the theoretical yield? d.If the actual yield is 386.0 g of CO2 , what isthe percent yield?

Explanation / Answer

345.0 g C2H6 * 1 mol C2H6/ 30.07 g = 11.47 mol C2H6 600.0 g O2 * 1 mol O2/ 32.0 g = 18.75 mol O2 2C2H6 + 7O2 >>>> 4CO2 + 6H2O   this is the balanced reaction (a) all C2H6 used 11.47 mol C2H6 * 4 mol CO2/ 2 mol C2H6 * 44.01 g CO2 / 1 mol =1009.6 f CO2 (b) limiting reagent for every 2 mol of C2H6 we need 7 mol of O2, if wemultiply 7/2 * 11.47 = 40.14 mol O2. and Clearly we do nothave enoough oxygen, so O2 is the limiting reagent (C) What is the theoretical yield? we use the limiting reagentwhich would run out first 18.75 mol O2 * 4 mol CO2 / 7 mol O2 * 44.01 g CO2/ 1 mol =471.5 g CO2 (d) Percebt yield % = (471.5-386.0)/ 386.0 * 100 = 22.15 %

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