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Luba_88 [7]
3 years ago
8

When ethyl acetoacetate (CH3COCH2CO2CH2CH3) is treated with one equivalent of CH3MgBr, a gas is evolved from the reaction mixtur

e, and after adding aqueous acid, ethyl acetoacetate is recovered in high yield. Identify the gas formed and explain why the starting material was recovered in this reaction. Be sure to answer all parts.

Chemistry
1 answer:
Alik [6]3 years ago
3 0

Answer:

Methane gas is formed in this reaction. After adding aqueous acid, ethyl acetoacetate is reformed through protonation of it.

Explanation:    

CH_{3}MgBr acts as base towards ethyl acetoacetate. Because ethyl acetoacetate contains active methylene group.

Hence  CH_{3}MgBr abstracts a proton from methylene group and being converted to methane gas.

After addition of aqueous acid, conjugate base of ethyl acetoacetate gets protonated and ethyl acetoacetate is reproduced.

See the attachment below.

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Answer:

4190.22 L = 4.19 m³.

Explanation:

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<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

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  • Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

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5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

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R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

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