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Standard XII

Chemistry

Percentage Composition

6 g urea samp...

Question

$6$ g urea sample on heating with $N a O H$ gives $0.224$ L $N H_{3}$ 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 :

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Solution

The molar mass of urea is $60$ g/mol.
$6$ g of urea corresponds to $\frac{6}{60} = 0.1$ moles.
They will produce $0.2$ moles of ammonia which will occupy a volume of $22.4 \times 0.2 = 4.48$ L.
Hence, the percentage yield is $\frac{0.224}{4.48} \times 100 = 5 %$ .

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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

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]

5.0 g

sample of urea when heated with

N a O B r

and

N a O H

gives

1120 m l

of nitrogen at STP. The percentage purity of the sample 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 :

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6 g urea sample on heating with NaOH gives 0.224 L NH3 at STP.NH2CONH2+2NaOH⟶2NH3+Na2CO3Thus, percentage yield of NH3 is :

The molar mass of urea is 60 g/mol.6 g of urea corresponds to 660=0.1 moles.They will produce 0.2 moles of ammonia which will occupy a volume of 22.4×0.2=4.48 L.Hence, the percentage yield is 0.2244.48×100=5%.

$6$ g urea sample on heating with $N a O H$ gives $0.224$ L $N H_{3}$ 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 :

The molar mass of urea is $60$ g/mol.
$6$ g of urea corresponds to $\frac{6}{60} = 0.1$ moles.
They will produce $0.2$ moles of ammonia which will occupy a volume of $22.4 \times 0.2 = 4.48$ L.
Hence, the percentage yield is $\frac{0.224}{4.48} \times 100 = 5 %$ .