Question

Calculate the theoretical percentage by weight of ethanol that is produced by fermentation of glucose.

Answer 1

\[C_{6}H_{12}0_{6}\rightarrow 2C_{2}H{5}OH + 2CO_{2}\]
So, In my book it says..

from the balanced equation, 1 mole of glucose (M = 180.2 g/mol) produces 2 moles (92.2g) of ethanol (M = 46.1 g/mol); in which I understand but..
" Thus, 100g of glucose produces 51.2g of ethanol i.e the theoretical yeild = 51.2%"

How did they able to calculate it to 51.2 g; i dont get it..

Answer 2

so we are just going to pick 100g of glucose as the starting material arbitrarily as it is easy to deal with 100g, we could pick any mass of glucose we would get the same answer.

1.

Convert to moles, mol = g/molecular mass

100g/180.2g/mol = 0.5549mol of glucose

2.

since it is 1 mol glucose for 2 mol ethanol we need to multiply the moles of glucose by 2
to get the moles of ethanol

\[ 0.5549mol_{glucose} * \frac{2mol_{ethanol}}{1mol_{glucose}} = 1.1098mol_{ethanol}\]

3.

1.1098mol = xg/46.1 g/mol
1.1098mol*46.1g/mol = 51.2g of ethanol

4.

the theoretical yield is thus
\[(\frac{Grams of Product}{GramsofReactant})*100\]

(51.2g/100g) = 51.2% yield

Answer 3

if you have any questions feel free to ask

Answer 4

i got it thanks.