Question

6 minutes ago • Chemistry • High School

Ethanol (C2H5OH) is produced from the fermentation of
sucrose in the presence of enzymes.
AC
C12H22011(aq) + H2O(g) 4 C2H5OH(1) + 4 CO2(g)
Determine the theoretical yield and the percent yields of ethanol if 720.g
sucrose undergoes fermentation and 323.0 g ethanol is obtained.

Answer 1

387.62, 83.33%

Step-by-step explanation:

Given:

`C_(12) H_(22) O_(11)(aq) + H_2O(g) -4 C_2H_5OH(1) + 4 CO_2(g)`

720g of Sucrose produces 323g ethanol.

Theoretical yield

To calculate the theoretical yield, you can use a mole-to-mole ratio. Assuming that there is excess H2O, you can calculate the theoretical yield like so:

`720g C_(12)H_(22)O_(11)*(1 mol C_(12)H_(22)O_(11))/(342.3g) *(4mol C_2H_5OH)/(1mol C_(12)H_(22)O_(11)) *(46.07g)/(1molC_2H_5OH) =387.62g`

Percent Yield

An easy way to find percent yield is

`(Actual )/(Theoretical) *100`

So, plug the numbers in.

`(323.0)/(387.62) *100 = 83.33%`

So, the percent yield is 83.33%.