Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes. C12H22O11(aq) + H2O(g) 4 C2H5OH(l) + 4 CO2(g)
Determine the theoretical yield and the percent yields of ethanol if 665 g sucrose undergoes fermentation and 305.0 g ethanol is obtained.
theoretical

actual yield of ethanol = 305.0g

molar mass of sucrose = 342g

molar mass of ethanol  =46g

mass of sucrose = 665g

mole of sucrose = mass / molar mass = 665/342

mole of sucrose =1.9 mole

sucrose :  C2H5OH

       1      :  4

     1.9     : 1.9x4 =7.6 mole of C2H5OH are formed

mass (therotical yield ) of C2H5OH= mole x mass

mass (therotical yield ) of C2H5OH= 7.6 x 46 = 349.6g

percent yields of ethanol =  actual /therotical x100

=305/349.6x100 = 87.24 %