A student carried out the fermentation of sucrose by using as starting material

Question

A student carried out the fermentation of sucrose by using as starting material 26.5 mL of a 35% sucrose solution (w:v).  How many mL of ethanol will be obtained if the enzymatic system works at 65% efficiency and the student gets a 70% recovery from the isolation procedure?

C12H22O11 + H2O rigtarrow Zymases 4 CH3CH2OH + 4 CO2 A student carried out the fermentation of sucrose by using as starting material 26.5 mL of a 35% sucrose solution (w:v).  How many mL of ethanol will be obtained if the enzymatic system works at 65% efficiency and the student gets a 70% recovery from the isolation procedure?

Explanation / Answer

Calculate mass of sucrose:
20.5mL * 35% (w:v) = 7.175 g

Calculate molecular weight of sucrose:
MW(C12H22O11) = 12*12.01 + 22*1.008 + 11*16.00 =
342.30 g/mol

Calculate moles of sucrose (n = m/MW):
n = 7.175/342.30 = 0.02096 mol

Due to stoichiometry the maximum amount of ethanol produced is 4 times the moles of sucrose. I.e. maximum yield is:
100% yield = 4*0.02096 = 0.08385 mol

Actual yield of ethanol (in moles) due to 65% efficiency (only 65% of sucrose is converted):
n(ethanol) = 65%*0.08385 = 0.05450 mol

So 0.05450 mol of ethanol is produced, but the student only recovers 70% of the product:
ethanol recovered = 70%*0.05450 = 0.03815 mol

Molecular weight of ethanol:
MW(CH3CH2OH) = 2*12.01 + 6*1.008 + 16.00 = 46.07 g/mol

mass of ethanol recovered (m = n*MW):
m = 0.03815*46.07 = 1.757 g

Look up the density of ethanol. From Wikipedia, density of ethanol at 25 deg C is 785.22 kg/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.