Which of the following on heating will not evolve N $_{2}$ gas?

_{3}

+ NaNO

_{2}

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_{4}

Cl + NaNO

_{2}

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_{2}

O + Cu

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(NH

_{4}

)

_{2}

_{2}

_{7}

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Solution

The correct option is B NH $_{3}$ + NaNO $_{2}$
The given reactions take place as:

$N H_{3} + N a N O_{2} \to N H_{3} N O_{2} + N a$

$N H_{4} C l + N a N O_{2} \to N_{2} + N a C l + 2 H_{2} O$

$N_{2} O + C u \to N_{2} + C u O$

$(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + H_{2} O$

Hence, the reaction between ammonia and sodium nitrate do not evolve $N_{2}$ gas.

Thus option A is the correct answer.

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Similar questions

N H_{3}

(N H_{4})_{2} S O_{4}, (N H_{4})_{2} C O_{3}, N H_{4} C l, N H_{4} N O_{3}, (N H_{4})_{2} C r_{2} O_{7}, (N H_{4})_{2} C_{2} O_{4}

(N H_{4})_{2} C r_{2} O_{7},

Which of the following will yield $N_{2}$ when heated?

Thermal decomposition of which of the following does/do not give $N_{2}$ gas?

I : {({N H}_{4})}_{2} {C r}_{2} O_{7} ⟶ N_{2} + {C r}_{2} O_{3} + H_{2} O

I I : {N H}_{4} {N O}_{3} ⟶ N_{2} O + 2 H_{2} O

I I I : 2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

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