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DerKrebs [107]
3 years ago
15

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o

f 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:
Chemistry
1 answer:
jekas [21]3 years ago
7 0

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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Calculate the percent yield when 500 grams of carbon dioxide react with an excess of water to produce 640 grams of carbonic acid
Elena-2011 [213]

Answer:

Percent yield =  90.5%

Explanation:

Given data:

Mass of carbon dioxide = 500 g

Mass of water = excess

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Percent yield = ?

Solution:

Balanced chemical equation:

CO₂ + H₂O  → H₂CO₃

Number of moles of carbon dioxide

Number of moles  = Mass / molar mass

Number of moles = 500 g/ 44 g/mol

Number of moles = 11.4 mol

Now we will compare the moles of H₂CO₃ with CO₂.

                              CO₂          :              H₂CO₃

                                 1             :                  1

                               11.4           :                11.4

Mass of carbonic acid:

Mass = number of moles × molar mass

Mass = 11.4 mol × 62.03 g/mol

Mass = 707.14 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

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Percent yield =  90.5%

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The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

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<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

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