What happens when ? (Give balanced equations)
(i)Sodium iodate is treated with sodium bisulphite solution.
(ii)Chlorine is passed through hot and concentrated NaOH solution.
(iii)Chlorine is passed into dilute and cold potassium hydroxide solution.
(iv)Chlorine gas is bubbled through a solution of ferrous bromide.
(v)Iodine reacts with concentrated `HNO_(3)`.
(vi)Chlorine is passed over slaked lime.
(vii)Potassium iodide is heated with `MnO_(2)` and concentrated `H_(2)SO_(4)`.
(viii)Chlorine reacts with `Na_(2)SO_(3)` solution.
(ix)Iodine is added to acidified stannous chloride solution.

Answer

Here is the step-by-step text solution for the given question along with hints for each step:

### (i) Sodium iodate is treated with sodium bisulphite solution.

**Step 1:** Write the reactants.
- Sodium iodate: NaIO₃
- Sodium bisulphite: NaHSO₃

**Step 2:** Identify the products.
- The reaction produces sodium bisulphate (NaHSO₄), sodium sulfate (Na₂SO₄), water (H₂O), and iodine (I₂).

**Step 3:** Write the unbalanced equation.
\[ \text{NaIO}_3 + \text{NaHSO}_3 \rightarrow \text{NaHSO}_4 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{I}_2 \]

**Step 4:** Balance the equation.
\[ 2 \text{NaIO}_3 + 5 \text{NaHSO}_3 \rightarrow 3 \text{NaHSO}_4 + 2 \text{Na}_2\text{SO}_4 + 3 \text{H}_2\text{O} + \text{I}_2 \]

### (ii) Chlorine is passed through hot and concentrated NaOH solution.

**Step 1:** Write the reactants.
- Chlorine: Cl₂
- Sodium hydroxide: NaOH (concentrated and hot)

**Step 2:** Identify the products.
- The reaction produces sodium chloride (NaCl), sodium chlorate (NaClO₃), and water (H₂O).

**Step 3:** Write the unbalanced equation.
\[ \text{Cl}_2 + \text{NaOH} \rightarrow \text{NaCl} + \text{NaClO}_3 + \text{H}_2\text{O} \]

**Step 4:** Balance the equation.
\[ 3 \text{Cl}_2 + 6 \text{NaOH} \rightarrow 5 \text{NaCl} + \text{NaClO}_3 + 3 \text{H}_2\text{O} \]

### (iii) Chlorine is passed into dilute and cold potassium hydroxide solution.

**Step 1:** Write the reactants.
- Chlorine: Cl₂
- Potassium hydroxide: KOH (dilute and cold)

**Step 2:** Identify the products.
- The reaction produces potassium chloride (KCl), potassium hypochlorite (KClO), and water (H₂O).

**Step 3:** Write the unbalanced equation.
\[ \text{Cl}_2 + \text{KOH} \rightarrow \text{KCl} + \text{KClO} + \text{H}_2\text{O} \]

**Step 4:** Balance the equation.
\[ \text{Cl}_2 + 2 \text{KOH} \rightarrow 2 \text{KCl} + \text{KClO} + \text{H}_2\text{O} \]

### (iv) Chlorine gas is bubbled through a solution of ferrous bromide.

**Step 1:** Write the reactants.
- Chlorine: Cl₂
- Ferrous bromide: FeBr₂

**Step 2:** Identify the products.
- The reaction produces ferric chloride (FeCl₃) and bromine (Br₂).

**Step 3:** Write the unbalanced equation.
\[ \text{Cl}_2 + \text{FeBr}_2 \rightarrow \text{FeCl}_3 + \text{Br}_2 \]

**Step 4:** Balance the equation.
\[ 3 \text{Cl}_2 + 2 \text{FeBr}_2 \rightarrow 2 \text{FeCl}_3 + 2 \text{Br}_2 \]

### (v) Iodine reacts with concentrated HNO₃.

**Step 1:** Write the reactants.
- Iodine: I₂
- Concentrated nitric acid: HNO₃

**Step 2:** Identify the products.
- The reaction produces iodic acid (HIO₃), nitrogen dioxide (NO₂), and water (H₂O).

**Step 3:** Write the unbalanced equation.
\[ \text{I}_2 + \text{HNO}_3 \rightarrow \text{HIO}_3 + \text{NO}_2 + \text{H}_2\text{O} \]

**Step 4:** Balance the equation.
\[ \text{I}_2 + 10 \text{HNO}_3 \rightarrow 2 \text{HIO}_3 + 10 \text{NO}_2 + 4 \text{H}_2\text{O} \]

### (vi) Chlorine is passed over slaked lime.

**Step 1:** Write the reactants.
- Chlorine: Cl₂
- Slaked lime: Ca(OH)₂

**Step 2:** Identify the products.
- The reaction produces bleaching powder (Ca(OCl)₂) and water (H₂O).

**Step 3:** Write the unbalanced equation.
\[ \text{Cl}_2 + \text{Ca(OH)}_2 \rightarrow \text{Ca(OCl)}_2 + \text{H}_2\text{O} \]

**Step 4:** Balance the equation.
\[ \text{Cl}_2 + \text{Ca(OH)}_2 \rightarrow \text{Ca(OCl)}_2 + 2 \text{H}_2\text{O} \]

### (vii) Potassium iodide is heated with MnO₂ and concentrated H₂SO₄.

**Step 1:** Write the reactants.
- Potassium iodide: KI
- Manganese dioxide: MnO₂
- Concentrated sulfuric acid: H₂SO₄

**Step 2:** Identify the products.
- The reaction produces potassium hydrogen sulfate (KHSO₄), manganese sulfate (MnSO₄), water (H₂O), and iodine (I₂).

**Step 3:** Write the unbalanced equation.
\[ \text{KI} + \text{MnO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{MnSO}_4 + \text{H}_2\text{O} + \text{I}_2 \]

**Step 4:** Balance the equation.
\[ 2 \text{KI} + \text{MnO}_2 + 3 \text{H}_2\text{SO}_4 \rightarrow 2 \text{KHSO}_4 + \text{MnSO}_4 + 2 \text{H}_2\text{O} + \text{I}_2 \]

### (viii) Chlorine reacts with Na₂SO₃ solution.

**Step 1:** Write the reactants.
- Chlorine: Cl₂
- Sodium sulfite: Na₂SO₃

**Step 2:** Identify the products.
- The reaction produces sodium bisulfate (NaHSO₄) and hydrochloric acid (HCl).

**Step 3:** Write the unbalanced equation.
\[ \text{Cl}_2 + \text{Na}_2\text{SO}_3 \rightarrow \text{NaHSO}_4 + \text{HCl} \]

**Step 4:** Balance the equation.
\[ 4 \text{Cl}_2 + \text{Na}_2\text{SO}_3 + 5 \text{H}_2\text{O} \rightarrow 2 \text{NaHSO}_4 + 8 \text{HCl} \]

### (ix) Iodine is added to acidified stannous chloride solution.

**Step 1:** Write the reactants.
- Iodine: I₂
- Stannous chloride: SnCl₂
- Acid (HCl)

**Step 2:** Identify the products.
- The reaction produces stannic chloride (SnCl₄) and hydroiodic acid (HI).

**Step 3:** Write the unbalanced equation.
\[ \text{I}_2 + \text{SnCl}_2 + \text{HCl} \rightarrow \text{SnCl}_4 + \text{HI} \]

**Step 4:** Balance the equation.
\[ \text{I}_2 + 2 \text{SnCl}_2 + 2 \text{HCl} \rightarrow 2 \text{SnCl}_4 + 2 \text{HI} \]

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