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Oxides of Nitrogen

A mixture of ...

Question

A mixture of ammonia and air at about $800^{0} C$ in the presence of $P t$ gauge forms:

N_{2} O

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N O

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N H_{2} O H

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N_{2} O_{2}

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Solution

The correct option is A $N O$
When ammonia is treated with air at 800 degree centigrade in presence platinum catalyst it produces water and $N O$ .
$4 N H_{3} + 5 O_{2} 800 d e g r e e ⟶ 4 N O + 6 H_{2} O$
Hence option B is correct.

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Similar questions

In the presence of platinum at $800^{0} C$ , ammonia reacts with oxygen to give X and water vapour. X then oxidizes to give reddish brown vapours of Y. What is X and Y?

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Q. Out of

N O, N O_{2}, N_{2} O, N_{2} O_{4}

and

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, the paramagnetic species are__________.

N H_{2} O H \to N_{2} O

In the above conversion, the equivalent weight of

N H_{2} O H

will be:

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In a mixture of

N_{2} O

and

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\frac{1}{3}

. The mole fraction of

N O

in a mixture of

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\frac{3}{4}

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molar ratio. the mole fraction of

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in the resulting mixture is:

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A mixture of ammonia and air at about 8000C in the presence of Pt gauge forms:

The correct option is A NOWhen ammonia is treated with air at 800 degree centigrade in presence platinum catalyst it produces water and NO.4NH3+5O2800degree⟶4NO+6H2OHence option B is correct.

A mixture of ammonia and air at about $800^{0} C$ in the presence of $P t$ gauge forms:

The correct option is A $N O$
When ammonia is treated with air at 800 degree centigrade in presence platinum catalyst it produces water and $N O$ .
$4 N H_{3} + 5 O_{2} 800 d e g r e e ⟶ 4 N O + 6 H_{2} O$
Hence option B is correct.