In the following reactions the sum of the total number of atoms in one molecule of A-B C isNaCl-K2Cr2O7- H2SO4-A-sideproducts0ex0exA-NaOH-B-sideproducts0ex0exB-H2SO4dil-H2O2-C-sideproducts

Identification of the reactants:NaClK_2Cr_2O_7/Conc. H_2SO_4→CrO_2Cl_2NaOH→Na_2CrO_4 + NaCl0ex0exNa_2CrO_4Dil. H_2SO_4→Na_2Cr_2O_7Dil. H_2O_2→CrO_50ex0ex(A)= Cr ...

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Standard XII

Chemistry

Curtius and Hofmann Rearrangements

In the follow...

Question

In the following reactions the sum of the total number of atoms in one molecule of (A),(B) & (C) is
$NaCl + K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4} \to (A) + side products (A) + NaOH \to (B) + side products (B) + H_{2} {SO}_{4} (dil .) + H_{2} O_{2} \to (C) + side products$

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Solution

Identification of the reactants:
$NaCl \overset{K_{2} {Cr}_{2} O_{7} / Conc . H_{2} {SO}_{4}}{\to} {CrO}_{2} {Cl}_{2} \overset{NaOH}{\to} {Na}_{2} {CrO}_{4} + NaCl {Na}_{2} {CrO}_{4} \overset{Dil . H_{2} {SO}_{4}}{\to} {Na}_{2} {Cr}_{2} O_{7} \overset{Dil . H_{2} O_{2}}{\to} {CrO}_{5} (A) = {CrO}_{2} {Cl}_{2}, (B) = {Na}_{2} {CrO}_{4} and (C) = {CrO}_{5}$
Therefore, the identified molecules are: $(A) = {CrO}_{2} {Cl}_{2}, (B) = {Na}_{2} {CrO}_{4} and (C) = {CrO}_{5}$ .

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In the following reactions the sum of the total number of atoms in one molecule of (A),(B) & (C) isNaCl+K2Cr2O7+H2SO4→(A)+sideproducts(A)+NaOH→(B)+sideproducts(B)+H2SO4(dil.)+H2O2→(C)+sideproducts

Identification of the reactants:NaCl→K2Cr2O7/Conc.H2SO4CrO2Cl2→NaOHNa2CrO4+NaClNa2CrO4→Dil.H2SO4Na2Cr2O7→Dil.H2O2CrO5(A)=CrO2Cl2,(B)=Na2CrO4and(C)=CrO5Therefore, the identified molecules are: (A)=CrO2Cl2,(B)=Na2CrO4and(C)=CrO5.

In the following reactions the sum of the total number of atoms in one molecule of (A),(B) & (C) is
$NaCl + K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4} \to (A) + side products (A) + NaOH \to (B) + side products (B) + H_{2} {SO}_{4} (dil .) + H_{2} O_{2} \to (C) + side products$