Question

C. When exposed to oxygen, fructose undergoes combustion. If you run the reaction with 10.0 g of fructose and 10.0 g of oxygen, how many moles of CO2 can you produce? The unbalanced equation for this reaction is: CeH12O6 + O2 → CO2 + H2O d. You run the reaction in lab and produce 9.49 g of CO,. Did the reaction go to completion? If not, what is your percent yield?

Answer 1

Ans. The reaction with fructose and oxygen can be written as Balanced equation - Co Higoo +60, 660 + 6H2O (exclushiesas 6x6I6X2) 6 (12+axlo 6 C 2*1+14) = 7&+ 12 +96 =192 gmax! 6 (12+32 6(18) = 180 g mol' 108 gmail 1. Gxuy =2649 moi!. produced 180 Fist find the limiting seagent. The amount'Ico, depend on limiting reagent. 1809 og Co the lo veact with 1928 og og 109 of 6Migos will heart with @ = 192x10 g og og R = 10067 g of a Only log of oz is given so oog is in limit amount. 19.29 of Og produces 264 g of coa 10g of Oz will produce coq = 264 xlo e of com = 264 x 10 g of Loa = 13.75 g of coa No. of moles (Og = Mass of coa = 13.75 g Molar Mass of log 44gmolt 0.3125 mol @ Theoretical yield of cong = 13.759 7. Experimental yield of code = 9.492

The Meaction is not completed as the theoretical yield is greater than the experimental yield. Percent yield of the reaction can be calculated as a xloo Porcent yield = Experimental yield - Theoretical Yield Percent yield - 9.49x100 13.75 I Percent yield = 69.01%] The percent yield of the reaction is 69.01%