if the percent yield for the following reaction is 65% how manygrams of KClo3 ar
ID: 691808 • Letter: I
Question
if the percent yield for the following reaction is 65% how manygrams of KClo3 are needed to produce 42.0 g of O2?Explanation / Answer
2 KCl O3 (a) - - - > 2 KCl(s) + 3 O2 (g) Molar Mass of KCl O3 = 39.098 +35.453 + ( 15.9994 x 3 ) = 122.5493 g / mole Molar Mass of O2 = ( 15.9994 x 2 ) = 31.9988 g / mole From balanced equation : ( 2 x 122.5493 ) g of KCl O3 is neededto give ( 3 x 31.9988 ) g of O2 [ Based on 100 % Theoretical Yield ] To obtain 42.0 g of O2 with 65.0 % Yield Amount of KCl O3 needed = 42.0 g of O2 x ( 245.0984 g of KCl O3/ 95.9964 g of O2 ) x ( 100 % / 65 % ) = 164.9763 g of KClO3 Answer : 165 g of KClO3
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