Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass of glucose is 180.15 g/mol, the molar mass of ethanol is 46.08 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol. What is the theoretical yield of ethanol from the fermentation of 61.5 g of glucose? theoretical yield: g If the reaction produced 23.4 g of ethanol, what is the percent yield? percent yield:

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-11-14T19:27:30+01:00

Answer:

The % yield is 74.45 %

Explanation:

Step 1: The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

Step 2: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

Step 3: Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

Step 4: Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

Step 5: Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

Step 6: Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %