Write a balanced chemical equation for the following reaction
${S O}_{2}$ gas is passed through an aqueous solution of a Fe(lll) salt.

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Solution

$S O_{2}$ acts as a reducing agent, when passed through an aqueous solution containing $F e (I I I)$ salt, it reduces $F e (I I I)$ to $F e (I I)$ , i.e., ferric $({F e}^{+ 3})$ ions to ferrous $({F e}^{+ 2})$ ions.
$2 {F e}^{3 +} + S O_{2} + 2 H_{2} O ⟶ 2 {F e}^{2 +} + {S O_{4}}^{2 -} + 4 H^{+}$

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Q. Question 20.3

Write the balanced chemical equation for the following reaction and identify the type of reaction.

Chlorine gas is passed in an aqueous potassium iodide solution to form potassium chloride solution and solid iodine.

Q. A red-brown metal X forms a salt XSO₄. When hydrogen sulphide gas is passed through an aqueous solution of XSO₄, then a black precipitate of XS is formed alongwith sulphuric acid solution.
(a) What could the salt XSO₄ be?
(b) What is the colour of salt XSO₄?
(c) Name the black precipitate XS.
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A red-brown metal X forms a salt $X S O_{4}$ . When hydrogen sulphide gas is passed through an aqueous solution of $X S O_{4}$ , then a black precipitate of XS is formed alongwith sulphuric acid solution.

(a) What could the salt $X S O_{4}$ be ?

(b) What is the colous of salt $X S O_{4}$ ?

(d) By using the formula of the salt obtained in (a) above, write an equation of the reaction which takes place when hydrogen sulphide gas is passed through its aqueous solution.

(e) What type of chemical reacion takes place in this case ?

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Write a balanced chemical equation for the following reactionSO2 gas is passed through an aqueous solution of a Fe(lll) salt.

SO2 acts as a reducing agent, when passed through an aqueous solution containing Fe(III) salt, it reduces Fe(III) to Fe(II), i.e., ferric (Fe+3) ions to ferrous (Fe+2) ions.2Fe3++SO2+2H2O⟶2Fe2++SO42−+4H+

Write a balanced chemical equation for the following reaction
${S O}_{2}$ gas is passed through an aqueous solution of a Fe(lll) salt.

$S O_{2}$ acts as a reducing agent, when passed through an aqueous solution containing $F e (I I I)$ salt, it reduces $F e (I I I)$ to $F e (I I)$ , i.e., ferric $({F e}^{+ 3})$ ions to ferrous $({F e}^{+ 2})$ ions.
$2 {F e}^{3 +} + S O_{2} + 2 H_{2} O ⟶ 2 {F e}^{2 +} + {S O_{4}}^{2 -} + 4 H^{+}$