Percent yield of NH3 in the following reaction is 80- NH2CONH2-2NaOH Na2CO3-2NH3 6 g NH2CONH2 reacts with 8 g NaOH to form NH3 equal to -

The correct option is B 2.72 g NH_2CONH_2+2NaOH Na_2CO_3+2NH_3 #160; #160; #160; #160; 6 g #160; #160; #160; #160; # ...

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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 $80 %$ .
$N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}$
$6$ g $N H_{2} C O N H_{2}$ reacts with $8$ g $N a O H$ to form $N H_{3}$ equal to :

3.4

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2.72

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4.25

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11.2

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Solution

The correct option is B $2.72$ g
$N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}$
$6$ g $8$ g
$\frac{6}{60} = 0.1$ mol $0.2$ mol
So, according to the reaction, $N H_{3}$ formed $= 0.2 \times 17 \times \frac{80}{100} = 2.72$ g

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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

x

g of

N H_{3}

. Find the value of

x

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}

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]

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 :

6

g of

H_{2}

reacts with

14

g of

N_{2}

to form

N H_{3}

till the reaction completely consumes the limiting reagent. The amount of another reactant in

g

left is:

Q. Hydrazine

(N_{2} H_{4})

is a rocket fuel which can be produced according to the following reaction:

C l N H_{2} + 2 N H_{3} \to N_{2} H_{4} + N H_{4} C l

When

1000

g of

C l N H_{2}

is reacted with excess of

N H_{3}

473

g of

N_{2} H_{4}

is produced. What is the percentage yield of the reaction?

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Percent yield of NH3 in the following reaction is 80%.NH2CONH2+2NaOHΔ−→Na2CO3+2NH36 g NH2CONH2 reacts with 8 g NaOH to form NH3 equal to :

The correct option is B 2.72 gNH2CONH2+2NaOHΔ−→Na2CO3+2NH3 6 g 8 g660=0.1 mol 0.2 mol So, according to the reaction, NH3 formed =0.2×17×80100=2.72 g

Percent yield of $N H_{3}$ in the following reaction is $80 %$ .
$N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}$
$6$ g $N H_{2} C O N H_{2}$ reacts with $8$ g $N a O H$ to form $N H_{3}$ equal to :

The correct option is B $2.72$ g
$N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}$
$6$ g $8$ g
$\frac{6}{60} = 0.1$ mol $0.2$ mol
So, according to the reaction, $N H_{3}$ formed $= 0.2 \times 17 \times \frac{80}{100} = 2.72$ g