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

JEE Advanced 2021 — Paper 2 — Question 36

Reaction of x gx\,\mathrm{g} of Sn\mathrm{Sn} with HCl\mathrm{HCl} quantitatively produced a salt. Entire amount of the salt reacted with y gy\,\mathrm{g} of nitrobenzene in the presence of required amount of HCl\mathrm{HCl} to produce 1.29 g1.29\,\mathrm{g} of an organic salt (quantitatively).

(Use molar masses in g mol−1\mathrm{g\,mol^{-1}} of H,C,N,O,Cl,Sn\mathrm{H,C,N,O,Cl,Sn} as 1,12,14,16,35,1191,12,14,16,35,119 respectively.)

The value of xx is ___\_\_\_ and the value of yy is ___\_\_\_.

The value of x\mathbf{x} is ______\_\_\_\_\_\_.

Answer: 3.57

Numerical answer — enter this value.

Step-by-step solution

Reduction reaction: C6H5NO2+3Sn+6HCl→C6H5NH3Cl+3SnCl2+2H2O\mathrm{C_6H_5NO_2 + 3Sn + 6HCl \rightarrow C_6H_5NH_3Cl + 3SnCl_2 + 2H_2O}

Molar mass of C6H5NH3Cl\mathrm{C_6H_5NH_3Cl}, M=6(12)+8(1)+14+35=129 g mol−1M = 6(12)+8(1)+14+35 = 129\,\mathrm{g\,mol^{-1}}

Given: n=1.29129=0.01 moln = \frac{1.29}{129} = 0.01\,\mathrm{mol}

Stoichiometry: 1 mol product→3 mol Sn1\,\mathrm{mol\ product} \rightarrow 3\,\mathrm{mol\ Sn}

Hence, nSn=3×0.01=0.03 moln_{\mathrm{Sn}} = 3 \times 0.01 = 0.03\,\mathrm{mol}

Molar mass of Sn\mathrm{Sn}, x=0.03×119=3.57 gx = 0.03 \times 119 = 3.57\,\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
Reaction of x\, g of Sn with HCl quantitatively produced a salt.… | JEE Advanced 2021 PYQ with Solution · DhiX AI