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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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olchik [2.2K]

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

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Ksp is defined as:

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<em>Where the concentrations [] are in equilibrium</em>

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are the actual concentrations</em>

<em />

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

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And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

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<em>Where X is defined as the reaction coordinate</em>

<em />

Replacing in Ksp expression:

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Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

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Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

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<h3>[Ag⁺] = 2.8x10⁻⁵M</h3>

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