Question

For $N{H_2}OH.HCl + NaN{O_2} \to \left( B \right)\xrightarrow{{Cu}}{\left( X \right)_{gas}}$ . Which of the following is correct?
A.(B) is an amphoteric oxide
B.(X) is a colourless, diamagnetic gas which combines with $Al$ on heating
C.(X) can be produced by action of $\left( {Zn + NaOH} \right)$ on $NaN{O_2}$
D.None of these

Answer 1

We have to know that the portrayal of synthetic reaction through images of substances as formulae is called compound condition. Adjusted chemical equation, a fair substance condition has number molecules of every component equivalent on both left and right sides of the reaction.

Complete answer:
We have to know that, the reaction somewhere in the range of $N{H_2}OH.HCl$ , and $NaN{O_2}$ gives nitrous oxide ( ${N_2}O$ ) as one of the items. It is then warmed to get dinitrogen gas within the sight of copper. The reaction is:
$N{H_2}OH.HCl + NaN{O_2} \to {N_2}O\xrightarrow[{Cu}]{\Delta }{N_2} \uparrow$
Where, B part is ${N_2}O$ and X part is ${N_2}$ .
For option (A),
The part B, ${N_2}O$ is a neutral oxide. Not an amphoteric oxide.
Therefore, option (A) is incorrect.
For option (B),
The part X is ${N_2}$ . This is a colourless gas. Then this gas is a diamagnetic in nature. This gas is combined with $Al$ on heating, it will give aluminum nitride. The reaction has to be given, $2Al + {N_2} \to 2AlN$
Therefore, the option (B) is correct.
For option (C), When zinc is reacted with sodium hydroxide on $NaN{O_2}$ to produce the part X is incorrect. Because it gives hydrogen gas.
Therefore, option (C) is incorrect.
Option (D) none of these is an incorrect option.

Hence, the correct option is (B).

Note:
To adjust the synthetic condition, you need to ensure the quantity of iotas of every component on the reactant side is equivalent to the quantity of molecules of every component on the item side. All together make the two sides equivalent, you should increase the quantity of particles in every component until the two sides are equivalent.