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JEE Advanced
Exam
JEE Advanced
Subject
Chemistry
Chapter
Practical Chemistry

JEE Advanced · Chemistry · Practical Chemistry

Paragraph: Metal nitrate (A) on heating decomposes, leaving a solid...

Paragraph:Metal nitrate (A)(A) on heating decomposes, leaving a solid residue (B)(B) which goes into solution with dilute HCl . The solution of (B) gives a white precipitate (C) with ammonium carbonate solution. The precipitate (C) is dissolved in dilute HCl and the solution is treated with potassium chromate to get yellow precipitate (D). The solution (B) with dilute H2SO4\mathrm{H}_2 \mathrm{SO}_4 also gives a white precipitate (E)(\mathrm{E}) insoluble in dilute HCl and nitric acid. The precipitate (E)(\mathrm{E}) is a part of a white pigment lithopone.Question:The compound (E)(\mathrm{E}) is :

Options

  1. ABaSO4\mathrm{BaSO}_4
  2. BMgSO4\mathrm{MgSO}_4
  3. CCaSO4\mathrm{CaSO}_4
  4. DNa2SO4\mathrm{Na}_2 \mathrm{SO}_4

Correct answer

A. BaSO4\mathrm{BaSO}_4

Step-by-step solution

Lithopone is a white pigment made from a mixture of ZnS and BaSO4Ba(NO3)2(A)→ΔBaO( B)+2NO2+12O2\mathrm{BaSO}_4\underset{(\mathrm{A})}{\mathrm{Ba}\left(\mathrm{NO}_3\right)_2} \xrightarrow{\Delta} \underset{(\mathrm{~B})}{\mathrm{BaO}}+2 \mathrm{NO}_2+\frac{1}{2} \mathrm{O}_2BaO+2HCl→BaCl2+H2OBaCl2+(NH4)2CO3→BaCO3↓(C)+2NH4ClBaCO3+2HCl→BaCl2+CO2+H2OBaCl2+K2CrO4→BaCrO4(D)+2KClBaCl2+H2SO4→BaSO4+2HCl\begin{aligned} & \mathrm{BaO}+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{BaCl}_2+\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3 \rightarrow \mathrm{BaCO}_3 \downarrow(\mathrm{C})+2 \mathrm{NH}_4 \mathrm{Cl} \\ & \mathrm{BaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{BaCl}_2+\mathrm{K}_2 \mathrm{CrO}_4 \rightarrow \mathrm{BaCrO}_4(\mathrm{D})+2 \mathrm{KCl} \\ & \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+2 \mathrm{HCl} \end{aligned}$$
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