Consider the following sequence of reactions. The percentage of...

The given starting material is diazoaminobenzene ($C_6H_5-N=N-NH-C_6H_5$).

When diazoaminobenzene is treated with an acid catalyst (HCl) and excess aniline, and warmed to $40-45^\circ \text{C}$, it undergoes a rearrangement reaction (or dissociation followed by coupling) to form $p$-aminoazobenzene.

The reaction sequence is as follows:
1. Diazoaminobenzene reacts with HCl to form benzenediazonium chloride and aniline.
2. The benzenediazonium ion then couples with the excess aniline at the para position to yield $p$-aminoazobenzene.
3. Cooling and adding acetic acid helps in precipitating the free base, $p$-aminoazobenzene, which is a yellow dye (often called Aniline Yellow).

The structure of the yellow product (X), $p$-aminoazobenzene, is $C_6H_5-N=N-C_6H_4-NH_2$.

The molecular formula of $p$-aminoazobenzene is $C_{12}H_{11}N_3$.

Let's calculate its molar mass:
Molar mass $= (12 \times 12) + (11 \times 1) + (3 \times 14)$
Molar mass $= 144 + 11 + 42 = 197 \text{ g mol}^{-1}$

The mass of nitrogen in one mole of the compound is $3 \times 14 = 42 \text{ g}$.

The percentage of nitrogen in the product is:
$\% \text{ N} = \dfrac{42}{197} \times 100 \approx 21.32\%$

Rounding to the nearest integer, we get $21$.

Answer: $21$