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3 years ago

N methyl 2 pyrrolidone heated with aqueous acid reaction

Chemistry

1 answer:

monitta3 years ago

5 0

<span>N-methyl 2 pyrrolidone is a cyclic amide and it when heated with aqueous acid gives the product as a cyclic amino acid. The product name is </span><span>N-Methyl-4-aminobutyric acid</span><span>

The reaction is given as follows in the attached picture.

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sergey [27]

Hello there!

I am not really sure which part you are talking about but let me answer some of them.

First, you need to read the information they gave you about glucose etc..

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I hope this helps!

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3 years ago

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elena55 [62]

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2 years ago

Read 2 more answers

Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$