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aliya0001 [1]
3 years ago
11

12dm3 ethene reacts with steam to make 18.4g ethanol.find the percentage yield

Chemistry
1 answer:
Veronika [31]3 years ago
3 0

% yield = 74.35

<h3>Further explanation</h3>

Given

12 dm³ ethene = 12 L

18.4 g ethanol(actual)

Required

The percentage yield

Solution

Reaction

C₂H₄(g) + H₂O(g) ⇒ C₂H₅OH(g)

Assume at STP, 1 mol gas = 22.4 L

mol ethene :

= 12 L : 22.4 L

= 0.538

From equation, mol ratio C₂H₅OH : C₂H₄ = 1 : 1, so mol C₂H₅OH = 0.538

Mass Ethanol (MW=46 g/mol) ⇒ theoretical

= 0.538 x 46

= 24.748 g

% yield = (actual/theoretical) x 100%

% yield = (18.4/24.748) x 100%

% yield = 74.35

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The partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.

Let x be the mole fraction of NO₂ and x' be the mole fraction of N₂O₄.

The total pressure according to Dalton's law of partial pressure is the sum of the partial pressures of each gas.

Let P be the total pressure of the gases, P' be the partial pressure of NO₂ = 1.4 bar and P" be the partial pressure of N₂O₄ = 0.46 bar.

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= 1.4 bar + 0.46 bar

= 1.86 bar

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= 1.4 bar/1.86 bar

= 0.753

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= 0.247

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So, PV = P₁V₁

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P₁ = (V/V₁)P

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P₁ = 0.93 bar

So, the <u>new</u> partial pressure of NO₂, P₂ = xP₁

= 0.753 × 0.93 bar

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= 0.23 bar

So, the partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.

Learn more about partial pressure here:

brainly.com/question/19813237

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