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dezoksy [38]
3 years ago
14

(25 hours ) /1 ( minutes) = ( hour )

Chemistry
1 answer:
Finger [1]3 years ago
3 0
60 minutes
thanks again I appreciate you I'm sorry but you ok if I invite to you in my prayers for you and Mani
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Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o
jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

<u>Step 1:</u> The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

<u>Step 2</u>: 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

<u>Step 3:</u> 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

<u>Step 4:</u> 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

<u>Step 5:</u> 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

<u>Step 6:</u> Calculate % yield

% yield = actual mass / theoretical mass

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

% yield = 74.45 %

The % yield is 74.45 %

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