1. Salt formed from tin and HCl

Tin reacts with hydrochloric acid to form stannous chloride:

$\text{Sn} + 2\text{HCl} \rightarrow \text{SnCl}_2 + \text{H}_2$

So, if tin is taken in amount $n$ mol, it produces $n$ mol of $\text{SnCl}_2$.

2. Reduction of nitrobenzene by $\text{SnCl}_2$/HCl

Nitrobenzene is reduced to aniline in acidic medium, and the aniline is obtained as aniline hydrochloride:

$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow[\text{HCl}]{\text{SnCl}_2} \text{C}_6\text{H}_5\text{NH}_2 \cdot \text{HCl}$

Oxidation state change of nitrogen:

• In $\text{NO}_2$: $+3$
• In $\text{NH}_2$: $-3$

Thus, reduction requires $6$ electrons per mole of nitrobenzene.

Now, $\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^-$ so 1 mol $\text{SnCl}_2$ supplies $2$ electrons.

Hence, to reduce 1 mol nitrobenzene, required $\text{SnCl}_2$ is:

$\frac{6}{2} = 3 \text{ mol}$

Therefore,

$3\text{SnCl}_2 + \text{C}_6\text{H}_5\text{NO}_2 + 7\text{HCl} \rightarrow \text{C}_6\text{H}_5\text{NH}_2\cdot\text{HCl} + 3\text{SnCl}_4 + 2\text{H}_2\text{O}$

So, $3 \text{ mol Sn} \rightarrow 1 \text{ mol aniline hydrochloride}$

3. Moles of organic salt formed

The organic salt is aniline hydrochloride:

$\text{C}_6\text{H}_5\text{NH}_2\cdot\text{HCl} = \text{C}_6\text{H}_8\text{NCl}$

Its molar mass is:

$6(12) + 8(1) + 14 + 35 = 72 + 8 + 14 + 35 = 129\,\text{g mol}^{-1}$

Given mass of salt formed = $1.29\,\text{g}$

So moles formed:

$n = \frac{1.29}{129} = 0.01\,\text{mol}$

4. Moles of tin required

From stoichiometry:

$3 \text{ mol Sn} \rightarrow 1 \text{ mol aniline hydrochloride}$

Therefore, for $0.01$ mol aniline hydrochloride:

$n_{\text{Sn}} = 3 \times 0.01 = 0.03\,\text{mol}$

Mass of tin:

$x = 0.03 \times 119 = 3.57\,\text{g}$

5. Final answer

$\boxed{x = 3.57\,\text{g}}$

This matches the stored correct answer.