Question

(i) A blue coloured compoud (A) on heating gives two products, (B) and (C).
(ii) A metal (D) is deposited on passing hydrogen through heated (B).
(iii) The solution of (B) in $HCl$ on treatment with gives a choclate brown coloured precipitated of compound (E).
(iv) (C) turns lime water milky which disappears on continous passage of (C) forming a compound (F).
Identify (A) to (F) and give chemical equations for the reactions at steps (i) to (iv).

• A. A is ${CuSO}_4{5H}_{2}O$
• B. B is $CuO$
• C. C is ${SO}_2$.
• D. F is ${Ca\left({HSO}_3\right)}_2$
• E. all of these correct

Answer 1

The correct option is E all of these correct

(i) A blue coloured compoud ${CuSO}_4{5H}_{2}O$ on heating gives two products, $CuO$ and ${SO}_2$.
${CuSO}_4{5H}_{2}O⟶CuO+{SO}_2+({5H}_{2}O+\frac{1}{2}{O}_2)$
(ii) A metal $Cu$ is deposited on passing hydrogen through heated $CuO$.
$CuO+H_2⟶Cu+H_{2}O$
(iii) The solution of $CuO$ in $HCl$ on treatment with gives a choclate brown coloured precipitated of compound ${Cu}_2\left[{Fe\left(CN\right)}_6\right]$.
$CuO+2HCl⟶{CuCl}_2+H_{2}O$
${2CuCl}_2+K_4\left[{Fe\left(CN\right)}_6\right]⟶{Cu}_2\left[{Fe\left(CN\right)}_6\right]+4KCl$
(iv) ${SO}_2$ turns lime water milky which disappears on continous passage of ${SO}_2$ forming a compound ${Ca\left({HSO}_3\right)}_2$
${Ca\left(OH\right)}_2+{SO}_2⟶\underset{milky}{{CaSO}_3}+H_{2}O$
${CaSO}_3+H_{2}O+{SO}_2⟶{Ca\left({HSO}_3\right)}_2$