Percent yield of NH3 in the following reaction is 80- NH2CONH2-2NaOH Na2CO3-2NH3- 6g of NH2CONH2 reacts with 8g NaOH to form xg of NH3- x is -Note- in g as nearest integer-

We have1 mole of urea is reacting with 2 mole of NaOH and give 2 mole of NH_3 So, moles of urea = 660 = 0.1 molSo, 0.1 mole of urea will produce 0.2 m ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

Percent yield...

Question

Percent yield of $N H_{3}$ in the following reaction is $80 % .$
$N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}$ .

6g of $N H_{2} C O N H_{2}$ reacts with 8g $N a O H$ to form $x$ g of $N H_{3}$ . $x$ is ____ .
[Note: in g as nearest integer]

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Solution

We have
1 mole of urea is reacting with 2 mole of NaOH and give 2 mole of $N H_{3}$

So, moles of urea = $\frac{6}{60} = 0.1$ mol

So, 0.1 mole of urea will produce 0.2 mole of $N H_{3}$

Moles of $N H_{3}$ = 0.2 mol
Mass of $N H_{3}$ produced = $0.2 * 17 = 3.4 g$

But the yield is 80%
So, amount of $N H_{3}$ produced = $\frac{3.4}{100} \times 80 = 2.7$ g

So, amount of $N H_{3}$ produced = 3g

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Similar questions

Q. Percent yield of

N H_{3}

in the following reaction is

80 %

N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}

6

N H_{2} C O N H_{2}

reacts with

8

N a O H

to form

N H_{3}

equal to :

Q. Percent yield of

N H_{3}

in the following reaction is

80 %

N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}

6

N H_{2} C O N H_{2}

reacts with

8

N a O H

to form

x

g of

N H_{3}

. Find the value of

x

6

g urea sample on heating with

N a O H

gives

0.224

N H_{3}

at STP.

N H_{2} C O N H_{2} + 2 N a O H ⟶ 2 N H_{3} + N a_{2} C O_{3}

Thus, percentage yield of

N H_{3}

is :

Q. Urea

(N H_{2} C O N H_{2})

is formed when

C O_{2}

reacts with

N H_{3}

C O_{2} + 2 N H_{3} ⟶ N H_{2} C O N H_{2} + H_{2} O

N H_{3}

and

C O_{2}

are obtained in the following steps:

I

N_{2} + 3 H_{2} 90 % - - \to 2 N H_{3}

I I

C a C O_{3} 50 % - - \to C a O + C O_{2}

Amount of

N_{2}

and

C a C O_{3}

required to produce

60

kg of urea are

X

kg and

Y

kg respectively. The value of

X + Y

(nearest integer value ) is :

Q. What will be the order of the reaction given below?

N H_{4} C N O \to N H_{2} C O N H_{2}

The reaction is completed in three steps as:

(i)

N H_{4} C N O ⇌ N H_{4} N C O

(fast equilibrium)

(ii)

N H_{4} N C O ⇌ N H_{3} + H - N = C = O

(fast equilibrium)

(iii)

N H_{3} + H - N = C = O \to N H_{2} C O N H_{2}

(slow equilibrium)

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Percent yield of NH3 in the following reaction is 80%. NH2CONH2+2NaOHΔ−→Na2CO3+2NH3. 6g of NH2CONH2 reacts with 8g NaOH to form xg of NH3. x is ____ .[Note: in g as nearest integer]

Percent yield of $N H_{3}$ in the following reaction is $80 % .$
$N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}$ .

6g of $N H_{2} C O N H_{2}$ reacts with 8g $N a O H$ to form $x$ g of $N H_{3}$ . $x$ is ____ .
[Note: in g as nearest integer]