I) Hydroxonium ion ii) Ammonium ion. 4. Calculate the volume of acidified potassium manganate(VII) required to react with 200 cm3 of 0.02 mol dm-3 hydrogen peroxide. The equation is 2MnO4-(aq) + 6H+(a

Here are the solutions to the questions:

1. i) Hydroxonium ion: $\boxed{\text{H}_3\text{O}^+}$ ii) Ammonium ion: $\boxed{\text{NH}_4^+}$

2. Step 1: Calculate the moles of hydrogen peroxide ($\text{H}_2\text{O}_2$).

   $$\text{Moles of H}_2\text{O}_2 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of H}_2\text{O}_2 = 0.02 \text{ mol dm}^{-3} \times 200 \text{ dm}^3 = 4 \text{ mol}$$

   Step 2: Use the mole ratio from the balanced ionic equation to find the moles of manganate(VII) ions ($\text{MnO}_4^-$).
   The equation is: $2\text{MnO}_4^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 5\text{H}_2\text{O}_2(\text{aq}) \longrightarrow 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l}) + 5\text{O}_2(\text{g})$
   From the equation, $5 \text{ moles of H}_2\text{O}_2$ react with $2 \text{ moles of MnO}_4^-$.

   $$\text{Moles of MnO}_4^- = \frac{2}{5} \times \text{Moles of H}_2\text{O}_2$$ $$\text{Moles of MnO}_4^- = \frac{2}{5} \times 4 \text{ mol} = 1.6 \text{ mol}$$

   Step 3: Calculate the volume of acidified potassium manganate(VII).

   $$\text{Volume of KMnO}_4 = \frac{\text{Moles of MnO}_4^-}{\text{Concentration of KMnO}_4}$$ $$\text{Volume of KMnO}_4 = \frac{1.6 \text{ mol}}{0.02 \text{ mol dm}^{-3}} = 80 \text{ dm}^3$$

   The volume of acidified potassium manganate(VII) is $\boxed{80 \text{ dm}^3}$.

3. Step 1: Calculate the moles of magnesium (Mg).

   $$\text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}$$ $$\text{Moles of Mg} = \frac{3.6 \text{ g}}{24 \text{ g/mol}} = 0.15 \text{ mol}$$

   Step 2: Calculate the mass of chlorine (Cl) in the chloride formed.

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