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An amorphous solid " $A$ " burns in air to form a gas " $B$ " which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $\mathrm{KMnO}_4$ solution and reduces $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$. Identify the solid " $A$ " and the gas " $B$ " and write the reactions involved.
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$\mathrm{SO}_2$ gas is formed as a by-product during roasting of sulphide ore. Therefore gas $\mathrm{B}$ is ' $\mathrm{SO}_2$ '. Since $\mathrm{SO}_2$ is formed when ' $\mathrm{S}_8$ ' is burnt in the air, therefore $\mathrm{A}$ is ' $\mathrm{S}_8{ }^2$ '. Reaction involved are given as :
(i) $\mathrm{S}_8+8 \mathrm{O}_2 \stackrel{\Delta}{\longrightarrow} 8 \mathrm{SO}_2$
(A)
(B)
(ii) $\mathrm{Ca}(\mathrm{OH})_2+\mathrm{SO}_2 \longrightarrow \mathrm{CaSO}_3+\mathrm{H}_2 \mathrm{O}$ lime water
$\mathrm{SO}_2$ turns lime water milky.
(iii) $\underset{\text { (Violet) }}{2 \mathrm{MnO}_4^{-}}+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow$
$5 \mathrm{SO}_4^{2-}+4 \mathrm{H}^{+}+\underset{(\text { Colourless })}{2 \mathrm{Mn}^{2+}}$
(iv) $2 \mathrm{Fe}^{3+}+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow$
$$
2 \mathrm{Fe}^{2+}+\mathrm{SO}_4^{2-}+4 \mathrm{H}^{+}
$$
(i) $\mathrm{S}_8+8 \mathrm{O}_2 \stackrel{\Delta}{\longrightarrow} 8 \mathrm{SO}_2$
(A)
(B)
(ii) $\mathrm{Ca}(\mathrm{OH})_2+\mathrm{SO}_2 \longrightarrow \mathrm{CaSO}_3+\mathrm{H}_2 \mathrm{O}$ lime water
$\mathrm{SO}_2$ turns lime water milky.
(iii) $\underset{\text { (Violet) }}{2 \mathrm{MnO}_4^{-}}+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow$
$5 \mathrm{SO}_4^{2-}+4 \mathrm{H}^{+}+\underset{(\text { Colourless })}{2 \mathrm{Mn}^{2+}}$
(iv) $2 \mathrm{Fe}^{3+}+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow$
$$
2 \mathrm{Fe}^{2+}+\mathrm{SO}_4^{2-}+4 \mathrm{H}^{+}
$$
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