The value of y is _________.

When metallic tin reacts quantitatively with hydrochloric acid, the only salt formed is $$SnCl_2$$:

$$Sn + 2\,HCl \;\rightarrow\; SnCl_2 + H_2$$

If the mass of tin taken is $$x\;{\rm g}$$, then the moles of $$SnCl_2$$ obtained are

$$n(SnCl_2)=\frac{x}{119}\qquad -(1)$$

Reduction of nitrobenzene to aniline in acid medium follows a 6-electron change. Each $$Sn^{2+}$$ $$\rightarrow$$ $$Sn^{4+}$$ supplies 2 electrons, so three moles of $$SnCl_2$$ are required per mole of nitrobenzene:

$$3\,SnCl_2 + C_6H_5NO_2 + 2\,HCl \;\longrightarrow\; 3\,SnCl_4 + C_6H_5NH_3Cl + 2\,H_2O$$

Hence moles of nitrobenzene that react are

$$n(C_6H_5NO_2)=\frac{n(SnCl_2)}{3}=\frac{x}{3\times119}\qquad -(2)$$

The organic salt obtained is anilinium chloride $$\left(C_6H_5NH_3Cl\right)$$. Its molar mass is

$$M\big(C_6H_5NH_3Cl\big)=6(12)+8(1)+14+35=129\;{\rm g\;mol^{-1}}$$

Given mass of this salt = $$1.29\;{\rm g}$$, the moles produced are

$$n\big(C_6H_5NH_3Cl\big)=\frac{1.29}{129}=0.01\;{\rm mol}\qquad -(3)$$

Because the salt forms quantitatively from nitrobenzene in a 1 : 1 ratio,

$$n(C_6H_5NO_2)=0.01\;{\rm mol}\qquad -(4)$$

Equating $$(2)$$ and $$(4)$$ gives

$$\frac{x}{3\times119}=0.01 \;\;\Longrightarrow\;\; x=0.01\times3\times119=3.57\;{\rm g}$$

Finally, the mass of nitrobenzene consumed (asked as $$y$$) is

$$y = n(C_6H_5NO_2)\times M(C_6H_5NO_2)$$

Molar mass of nitrobenzene $$C_6H_5NO_2$$:

$$M(C_6H_5NO_2)=6(12)+5(1)+14+2(16)=123\;{\rm g\;mol^{-1}}$$

Therefore

$$y = 0.01 \times 123 = 1.23\;{\rm g}$$

So, the value of $$y$$ is 1.23 g.

Answer: 1.23 - 1.23