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Compounds Containing Nitrogen question

2021 · Shift 2 · Q10
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Compounds Containing Nitrogen question

2021 · Shift 2 · Q10

JEE AdvancedChemistryCompounds Containing NitrogenNumerical+2 / −1
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with 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) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively). The value of y is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1.23

  1. Salt formed from tin and HCl

Tin reacts with hydrochloric acid to give stannous chloride:

Sn+2HCl→SnCl2+H2\text{Sn} + 2\text{HCl} \rightarrow \text{SnCl}_2 + \text{H}_2Sn+2HCl→SnCl2​+H2​

So, the salt formed is SnCl2\text{SnCl}_2SnCl2​.

  1. Reduction of nitrobenzene by SnCl2\text{SnCl}_2SnCl2​/HCl

Nitrobenzene is reduced to aniline in acidic medium, and aniline in presence of HCl forms anilinium chloride:

C6H5NO2→HClSnCl2C6H5NH2→HClC6H5NH3Cl\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow[\text{HCl}]{\text{SnCl}_2} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{HCl}} \text{C}_6\text{H}_5\text{NH}_3\text{Cl}C6​H5​NO2​SnCl2​HCl​C6​H5​NH2​HCl​C6​H5​NH3​Cl

The final organic salt is anilinium chloride, C6H5NH3Cl\text{C}_6\text{H}_5\text{NH}_3\text{Cl}C6​H5​NH3​Cl.

  1. Molar mass of anilinium chloride

For C6H5NH3Cl\text{C}_6\text{H}_5\text{NH}_3\text{Cl}C6​H5​NH3​Cl, total formula is C6H8NCl\text{C}_6\text{H}_8\text{NCl}C6​H8​NCl.

M=6(12)+8(1)+14+35=72+8+14+35=129M = 6(12) + 8(1) + 14 + 35 = 72 + 8 + 14 + 35 = 129M=6(12)+8(1)+14+35=72+8+14+35=129

Given mass of organic salt = 1.29 g1.29\,\text{g}1.29g

n(anilinium chloride)=1.29129=0.01 moln(\text{anilinium chloride}) = \frac{1.29}{129} = 0.01\,\text{mol}n(anilinium chloride)=1291.29​=0.01mol

Thus, moles of nitrobenzene used = 0.010.010.01 mol, since 1 mol nitrobenzene gives 1 mol anilinium chloride.

  1. Molar mass of nitrobenzene

Nitrobenzene = C6H5NO2\text{C}_6\text{H}_5\text{NO}_2C6​H5​NO2​

M=6(12)+5(1)+14+2(16)=72+5+14+32=123M = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123M=6(12)+5(1)+14+2(16)=72+5+14+32=123

Therefore,

y=0.01×123=1.23 gy = 0.01 \times 123 = 1.23\,\text{g}y=0.01×123=1.23g

  1. Final answer

1.23\boxed{1.23}1.23​

This matches the stored correct answer.

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