1) Calculate the percent by volume of acetic acid by fermentation. Assumptions: No volume changed during fermentation. Overall chemical reaction facilitated by Acetobacter aceti: C2H5OH O2 CH3COOH H2O 6% Given: Volume of 6% C2H5OH = 3 L Molar mass of C2H5OH = 46.07 g/mol Molar mass of CH3COOH = 60.06 g/mol Density of C2H5OH = 0.789 g/mL Density of CH3COOH = 1.05 g/mL Required: Percent acetic (% by volume) produced

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1) Calculate the percent by volume of acetic acid by fermentation. Assumptions: No volume
changed during fermentation.
Overall chemical reaction facilitated by Acetobacter aceti:
C2H5OH
+
O2
CH3COOH
H20
6%
Given: Volume of 6% C2H5OH = 3 L
Molar mass of C2H5OH = 46.07 g/mol
Molar mass of CH3COOH = 60.06 g/mol
Density of C2H5OH = 0.789 g/mL
Density of CH3COOH = 1.05 g/mL
Required: Percent acetic (% by volume) produced
Transcribed Image Text:1) Calculate the percent by volume of acetic acid by fermentation. Assumptions: No volume changed during fermentation. Overall chemical reaction facilitated by Acetobacter aceti: C2H5OH + O2 CH3COOH H20 6% Given: Volume of 6% C2H5OH = 3 L Molar mass of C2H5OH = 46.07 g/mol Molar mass of CH3COOH = 60.06 g/mol Density of C2H5OH = 0.789 g/mL Density of CH3COOH = 1.05 g/mL Required: Percent acetic (% by volume) produced
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