what mass of ethanol could form from fermentation of 90.0g of glucose? what is the percentage yield if 6.0g of ethanol forms?

Answer 1

C6H12O6 ==> 2C2H5OH + 2CO2

mols glucose = grams/molar mass = estimated 0.5
Using the coefficients in the balanced equation, convert mols glucose to mols ethanol. That is 0.5 x (2 mols ethanol/1 mol glucose) = 0.5 x 2 = 1 mol ethanol produced.
Then g ethanol = mols x molar mass = estimated 1 mol x 46g/mol = about 46 grams. That is the theoretical yield (TY). The actual yield (AY) is 6.0g
%yield = (AY/TY)*100 = ?
You need to go through and clear up the numbers.