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Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes.
C12H22O11(aq) + H2O(g) 4 C2H5OH(l) + 4 CO2(g)
Determine the theoretical yield and the percent yields of ethanol if 735 g sucrose undergoes fermentation and 310.5 g ethanol is obtained.
theoretical
g
percent
%

Ethanol C2H5OH is produced from the fermentation of sucrose in the presence of enzymes C12H22O11aq H2Og 4 C2H5OHl 4 CO2g Determine the theoretical yield and the class=

Respuesta :

serandulnathsm

Explanation:

to calculate theoratical yield...first we must calculate how much Sucrose moles we have..in order to to do so...we must calculate molar mass of sucrose.

molar mass of sucrose is 342.3 g/mol.

now we can calculate how many sucrose moles we have by dividing the mass with molar mass of sucrose

735g/(342.3g/mol)=2.147mol

theoratically...according to stoichiometry..every 1 mole of sucrose yields 4 moles of ethanol...so...

2.147mole yield 2.147*4mol=8.588mol

now we must calculate weight of that much ethanol..molar mass of ethanol is 46.07 g/mol...

so we can multiple moles by molar mass to obtain the weight 8.588mol*46.07g/mol=395.649g

but we only obtained 310.5g...so percentage we have is

[tex] \frac{310.5}{395.649} \times 100[/tex]

78.47%

if there is trouble with molar mases I used....use what you calculated...

If my explanation is good enough...please mark it as brainliest.thanks