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Question

From the given series of reactions, select the correct option(s):
(I)NH3+O2NO+H2O
(II)NO+O2NO2
(III)NO2+H2OHNO3+HNO2
(IV)HNO2HNO3+NO+H2O

A
Moles of HNO3 obtained is half the number of moles of ammonia used if HNO2 is not used to produce HNO3 by reaction (IV)
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B
1006 % more HNO3 will be produced if HNO2 is used to produce HNO3 by reaction (IV) than if HNO2 is not used to produce HNO3 by reaction (IV)
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C
If HNO2 is used to produce HNO3 then 14th of total HNO3 is produced by reaction (IV)
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D
Moles of NO produced in reaction (IV) is 50% of moles of total HNO3 produced.
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Solution

The correct options are
A Moles of HNO3 obtained is half the number of moles of ammonia used if HNO2 is not used to produce HNO3 by reaction (IV)
C If HNO2 is used to produce HNO3 then 14th of total HNO3 is produced by reaction (IV)
D Moles of NO produced in reaction (IV) is 50% of moles of total HNO3 produced.
The balanced chemical equations are:

(I) 4NH3+5O24NO+6H2O
(II) 2NO+O22NO2
(III) 2NO2+H2OHNO3+HNO2
(IV) 3HNO2HNO3+2NO+H2O

Let the moles of NH3 be x moles.

moles of HNO3 in III reaction=x×44×22×12=x2

moles of HNO3 in IV reaction=x×44×22×12×13=x6

Moles of NO=x×44×22×12×23=x3

Total moles of HNO3=x2+x6=2x3

A) Moles of HNO3 obtained in III reaction if HNO2 is not used to produce HNO3 = x2
=NH32

B) The amount of HNO3 produced,
=x6x2×100

=1003% of HNO3 in IV reaction.

C) The amount of HNO3 produced in IV reaction,
=x62x3=14 of the total amount of HNO3 produced.

D) moles of NOTotal moles of HNO3×100=x32x3×100

= 50 %

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