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On heating ammonium dichromate and barium azide separately we get

(i) $N_{2}$ in both cases

(ii) $N_{2}$ with ammonium dichromate and $NO$ with barium azide

(iii) $N_{2}O$ with ammonium dichromate and $N_{2}$ with barium azide

(iv) $N_{2}O$ with ammonium dichromate and $NO_{2}$ with barium azide

Answers (1)

The answer is the option (i)

$\left ( NH_{4} \right )_{2}Cr_{2}O_{7}\rightarrow N_{2}+Cr_{2}O_{3}+4H_{2}O$

$Ba \left ( N_{3} \right )_{2}\rightarrow 3N_{2}+Ba$

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infoexpert23

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