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,Sn as 1,12,14,16,35,119 respectively.) The value of x is \ \ \ and the value of y is \ \ \ . The value of y is \ \ \ \ \ .

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Advanced 2021 — Paper 2 — Question 37

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

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

The value of $x$ is $\_\_\_$ and the value of $y$ is $\_\_\_$ .

The value of $\mathbf{y}$ is $\_\_\_\_\_$ .

Answer: 1.23

Numerical answer — enter this value.

Step-by-step solution

Same reduction reaction: $\mathrm{C_6H_5NO_2 + 3Sn + 6HCl \rightarrow C_6H_5NH_3Cl + 3SnCl_2 + 2H_2O}$

From above, we have

n_{\mathrm{product}} = 0.01\,\mathrm{mol}

Thus, $n_{\mathrm{nitrobenzene}} = 0.01\,\mathrm{mol}$

Molar mass of $\mathrm{C_6H_5NO_2}$ , $M = 6(12) + 5(1) + 14 + 2(16) = 123\,\mathrm{g\,mol^{-1}}$

Therefore, $y = 0.01 \times 123 = 1.23\,\mathrm{g}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Chemical Equations, Stoichiometry and Limiting Reagent

Answer: 1.23

Step-by-step solution

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