Question

Calculate the percentage yield of ethanol when 100 g of glucose is fermented to make 20 g of ethanol.

Answer 1

the normal reaction of glucose fermentation to ethanol is

$C_6H_{12}O_{6(aq)}\to 2C_2H_5OH_{(aq)} +2CO_{2(g)}$

from the reaction above,

1 mole of glucose ferments to form 2 moles of ethanol

This means that, 180g of glucose ferments to form 92g (2×46) of ethanol

100g of glucose should therefore ferment to form 51.1g ($\dfrac{100×92}{180}$) of ethanol

from, $\% yield = \dfrac{\textsf{Actual Yield}}{\textsf{Theoretical Yield}} × 100\%$

$\%yield = \dfrac{20g}{51.1g} × 100\% = 39.14$%

This means that the percentage yield of ethanol when 100 g of glucose is fermented is 39.14%