Answered

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 :

purekainex

Answer:

387.62, 83.33%

Explanation:

Given:

[tex]C_{12} H_{22} O_{11}(aq) + H_2O(g) -4 C_2H_5OH(1) + 4 CO_2(g)[/tex]

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:

[tex]720g C_{12}H_{22}O_{11}*\frac{1 mol C_{12}H_{22}O_{11}}{342.3g} *\frac{4mol C_2H_5OH}{1mol C_{12}H_{22}O_{11}} *\frac{46.07g}{1molC_2H_5OH} =387.62g[/tex]

Percent Yield

An easy way to find percent yield is

[tex]\frac{Actual }{Theoretical} *100[/tex]

So, plug the numbers in.

[tex]\frac{323.0}{387.62} *100 = 83.33%[/tex]

So, the percent yield is 83.33%.

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