Glucose (C6H12O6)(C6H12O6) can be fermented to yield ethanol (CH3CH2OH)(CH3CH2OH) and carbon dioxide (CO2).

C6H12O6⟶2CH3CH2OH+2CO2

The molar mass of glucose is 180.15 g/mol,180.15 g/mol, the molar mass of ethanol is 46.08 g/mol,46.08 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol.

a) What is the theoretical yield (in grams) of ethanol from the fermentation of 97.5 g of glucose?

b) If the reaction produced 23.4 g of ethanol, what was the percent yield?

a) 49.8 gram

b) 47.0 %

Explanation:

First we have to calculate the moles of glucose

The balanced chemical reaction will be,

From the balanced reaction, we conclude that

As,1 mole of glucose produce = 2 moles of ethanol

So, 0.54 moles of glucose will produce =   mole of ethanol

Now we have to calculate the mass of ethanol produced

Now we have to calculate the percent yield of ethanol

Therefore, the percent yield is 47.0 %

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)