### Step 1: Writing the Balanced Chemical Equation
In an acidic medium, the reaction between chlorate ions (ClO3-) and sulfur dioxide (SO2) to produce sulfate ions (SO42-) and chloride ions (Cl-) can be balanced as follows:
1. Write the unbalanced equation:
[tex]\[
\text{ClO}_3^- + \text{SO}_2 \rightarrow \text{SO}_4^{2-} + \text{Cl}^-
\][/tex]
2. Identify oxidation states and balance the redox reaction:
- Oxidation half-reaction:
[tex]\[
\text{SO}_2 \rightarrow \text{SO}_4^{2-} + 2\text{e}^-
\][/tex]
- Reduction half-reaction:
[tex]\[
\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}
\][/tex]
3. Balance the electrons in the half-reactions:
[tex]\[
\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}
\][/tex]
[tex]\[
3\text{SO}_2 \rightarrow 3\text{SO}_4^{2-} + 6\text{e}^-
\][/tex]
4. Combine the balanced half-reactions (ensure electrons cancel):
[tex]\[
\text{ClO}_3^- + 3\text{SO}_2 + 6\text{H}^+ \rightarrow 3\text{SO}_4^{2-} + \text{Cl}^- + 3\text{H}_2\text{O}
\][/tex]
5. Simplify if necessary:
[tex]\[
\text{ClO}_3^- + 3\text{SO}_2 + \text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 2\text{H}^+ + \text{Cl}^-
\][/tex]
Therefore, the balanced equation in an acidic medium is:
[tex]\[
\text{ClO}_3^- + 3\text{SO}_2 + \text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 2\text{H}^+ + \text{Cl}^-
\][/tex]
### Step 2: Calculating the Mass of Citric Acid Produced
We need to find the mass of citric acid (C6H8O7) produced from exactly 1 metric ton (1000 × 103 kg).
1. Understand the relation and conversion factors:
- Molecular weight of citric acid (C6H8O7) is approximately 192.12 g/mol.
2. Convert metric ton to grams:
[tex]\[ 1 \text{ metric ton} = 1000 \times 10^3 \text{ kg} = 10^6 \text{ g} \][/tex]
3. Calculate the moles of citric acid produced:
[tex]\[
\text{Moles of citric acid} = \frac{10^6 \text{ g}}{192.12 \text{ g/mol}} \approx 5205.08 \text{ mol}
\][/tex]
Thus, the mass of citric acid produced from exactly 1 metric ton is:
[tex]\[
\approx 5205.08 \text{ mol}
\][/tex]
This provides us with a clear and detailed solution for the given problem.
