Inthereactionshown,100.0gC6H11OHyielded64.0g C6H10 . (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of C6H11OH would produce 100.0 g C6H10 if the percent yield is that determined in part (b)?
C6H11OH ¡ C6H10 + H2O
Inthereactionshown,100.0gC6H11OHyielded64.0g C6H10 . (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of C6H11OH would produce 100.0 g C6H10 if the percent yield is that determined in part (b)?
C6H11OH ¡ C6H10 + H2O
2 answers
C6H11OH ==> C6H10 + H2O
Theoretical yield for 100.0 g will produce 82 g C6H10 so TY = 82 grams.
%yield = (64/82)*100 = approx 78%.
C) You want 100.0 g C6H10. mols C6H10 in 100 g = 100/about 1.2 mols so you would need 1.2 mols to start if the reaction were 100%. It isn't so you must start with 1.2/0.78 = about 1.6 mols C6H11OH. Then grams C6H11OH = mols x molar mass = ?
Theoretical yield for 100.0 g will produce 82 g C6H10 so TY = 82 grams.
%yield = (64/82)*100 = approx 78%.
C) You want 100.0 g C6H10. mols C6H10 in 100 g = 100/about 1.2 mols so you would need 1.2 mols to start if the reaction were 100%. It isn't so you must start with 1.2/0.78 = about 1.6 mols C6H11OH. Then grams C6H11OH = mols x molar mass = ?