Question

A white crystalline solid P on strong heating decomposes to give a reddish brown gas R and residue Q ,which is yellow when hot and white when cold. The solution of P gives a white precipitate S with sodium hydroxide solution which dissolves in excess sodium hydroxide .
i) Identify P,Q,R,S .
ii) Give balanced equation for each: 1) Thermal decomposition of P.
2) S dissolves in excess NaOH
iii) What is your observation when solution of P is mixed with Ferrous sulphate solution followed by few drops of concentrated H2SO4?

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Answer 1

I P=Zn(NO³)²
R=NO²
Q=ZnO
S=Zn(OH)²

ii I) Zn(NO³)²=ZnO+2NO²+O²
iii NaOH + Zn(OH)²== Na²ZnO²+H²O

3 On mixing P with ferrous sulphate solution followed by feeding drops of conc. H²So⁴,a brown ring will develop at the junction of H²So⁴ and P.

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Answer 2

Answer:

In the given data, P is zinc nitrate Zn(NO₃)₂, Q is zinc oxide ZnO, R is nitrogen dioxide NO₂ and S is Zinc hydroxide Zn(OH)₂.

Explanation:

Brown ring test is performed to find the presence of nitrate ions.

When nitrate ions mixed with ferrous sulphate followed by few drops of conc. sulphuric acid. A brown ring will appear on the junction of two rings. It means solid P is any nitrate.

Zinc oxide ZnO which is yellow when hot and white when cold.

Therefore, the nitrate is actually zinc nitrate. So solid P is zinc nitrate.

• Now, thermal decomposition of zinc nitrate:

             $2Zn(NO_{3})_{2}$   →    $2ZnO$   $+$  $4NO_{2}$   $+$   $O_{2}$

       Therefore,  reddish brown gas R is nitrogen dioxide and residue Q   is zinc oxide.

• $Zn(NO_{3})_{2}$  reaction with sodium hydroxide:

              $Zn(NO_{3})_{2}$   $+$   $2NaOH$   →    $Zn(OH)_{2}$  $+$   $2NaNO_{3}$

        The white precipitate of S is zinc hydroxide Zn(OH)₂.

• $Zn(OH)_{2}$  reaction with excess of sodium hydroxide gives colorless solution.

             $Zn(OH)_{2}$  $+$   $2NaOH$   →  $Na_{2}ZnO_{2}$   +  $2H_{2}O$