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JEE Advanced 2021 · Paper 2, Chemistry question 9

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Question Stem for Question Nos. 9 and 10

Reaction of x\mathbf x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y\mathbf y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).

(Use Molar masses (in g mol−1^{-1}) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).

The value of x\mathbf x is ____.

Answer and worked solution

Official answer: 3.57

The reactions.

  • Sn + 2HCl gives SnClX2\ce{SnCl2}.
  • SnClX2\ce{SnCl2}/HCl reduces nitrobenzene to anilinium chloride:
    CX6HX5NOX2+3 SnClX2+7 HCl→CX6HX5NHX3Cl+3 SnClX4+2 HX2O.\ce{C6H5NO2 + 3SnCl2 + 7HCl -> C6H5NH3Cl + 3SnCl4 + 2H2O}.

Moles. Anilinium chloride has M=129M=129, so 1.29129=0.01\frac{1.29}{129}=0.01 mol. This needs 3×0.01=0.033\times0.01=0.03 mol of SnClX2\ce{SnCl2}, i.e. 0.03 mol of Sn.

Mass of Sn. x=0.03×119=3.57x=0.03\times119=3.57 g.

Topics:StoichiometryNitro compounds
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