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Question

# From the given series of reactions, select the correct option(s): (I)NH3+O2→NO+H2O (II)NO+O2→NO2 (III)NO2+H2O→HNO3+HNO2 (IV)HNO2→HNO3+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+5O2→4NO+6H2O (II) 2NO+O2→2NO2 (III) 2NO2+H2O→HNO3+HNO2 (IV) 3HNO2→HNO3+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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