An orange solid (X) on heating gives a colour | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The thermal decomposition of ammonium dichromate is given by: $(NH_4)_2Cr_2O_7 \rightarrow N_2(g) + Cr_2O_3(s) + 4H_2O(g)$.Here,$(X)$ is $(NH_4)_2

Vedclass · Accepted Answer

$Mg_3N_2$

Vedclass · Answer

(A) The thermal decomposition of ammonium dichromate is given by: $(NH_4)_2Cr_2O_7 \rightarrow N_2(g) + Cr_2O_3(s) + 4H_2O(g)$.Here,$(X)$ is $(NH_4)_2Cr_2O_7$ (orange solid),$(Y)$ is $N_2$ (colourless gas),and $(Z)$ is $Cr_2O_3$ (green residue).When the gas $(Y)$,which is $N_2$,reacts with magnesium metal,it forms magnesium nitride:$3Mg + N_2 \rightarrow Mg_3N_2$.$Mg_3N_2$ is a white solid substance.

An orange solid $(X)$ on heating gives a colourless gas $(Y)$ and only a green residue $(Z)$. Gas $(Y)$ on treatment with $Mg$ produces a white solid substance. Identify the white solid substance.

Similar Questions

Which halide of nitrogen is least basic?

What are the products formed when ammonia reacts with excess chlorine?

Assertion $(A)$: $P_4O_{10}$ cannot be used to remove moisture from ammonia gas. Reason $(R)$: $P_4O_{10}$ reacts with $NH_3$ gas. The correct answer is:

The number of $P-O-P$ bonds in $H_4P_2O_7$,$(HPO_3)_3$ and $P_4O_{10}$ are respectively.

$N_2$ combines with metals to form:

Vedclass Test Series

Exam Paper Generator

Online Exam Module