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Yes, you are correct: 1 ton is equivalent to 1000 kg.

Now, let's solve the mass balance problem step by step.

### Problem Breakdown:
1. **NaCl used**: 95% purity
2. **H₂SO₄ used**: 92.5% purity
3. **Decomposition degree of NaCl**: 92%
4. **Product**: 1 ton (1000 kg) of sodium sulfate (Na₂SO₄)

### Reaction:

The reaction between sodium chloride (NaCl) and sulfuric acid (H₂SO₄) to form sodium sulfate and hydrochloric acid (HCl) is:
\[
2 \, \text{NaCl} + \text{H₂SO₄} \rightarrow \text{Na₂SO₄} + 2 \, \text{HCl}
\]

We will calculate how much NaCl and H₂SO₄ are needed to produce 1000 kg of Na₂SO₄.

### Step 1: Molar Mass Calculation
- Molar mass of NaCl = 58.44 g/mol
- Molar mass of H₂SO₄ = 98.08 g/mol
- Molar mass of Na₂SO₄ = 142.04 g/mol

### Step 2: Stoichiometry of the Reaction
According to the equation, 2 moles of NaCl react with 1 mole of H₂SO₄ to produce 1 mole of Na₂SO₄. This means for every 2 moles of NaCl, we get 1 mole of Na₂SO₄.

### Step 3: Calculate Required Na₂SO₄
You need 1 ton (1000 kg) of Na₂SO₄, so:
\[
\text{Moles of Na₂SO₄} = \frac{1000 \, \text{kg}}{142.04 \, \text{g/mol}} = 7040.5 \, \text{mol}
\]

### Step 4: Calculate Required NaCl
From the stoichiometry, 1 mole of Na₂SO₄ requires 2 moles of NaCl. Therefore, the number of moles of NaCl required is:
\[
\text{Moles of NaCl} = 2 \times 7040.5 = 14081 \, \text{mol}
\]
Now, convert moles to mass:
\[
\text{Mass of NaCl} = 14081 \times 58.44 \, \text{g/mol} = 822876 \, \text{g} = 822.88 \, \text{kg}
\]

Since the NaCl is 95% pure, the actual amount of NaCl needed is:
\[
\text{Mass of NaCl} = \frac{822.88}{0.95} = 866.19 \, \text{kg}
\]

Taking into account that only 92% of the NaCl decomposes, the required mass is:
\[
\text{Adjusted mass of NaCl} = \frac{866.19}{0.92} = 941.51 \, \text{kg}
\]

### Step 5: Calculate Required H₂SO₄
According to the reaction, 1 mole of Na₂SO₄ requires 1 mole of H₂SO₄. Therefore, the number of moles of H₂SO₄ needed is the same as the number of moles of Na₂SO₄ (7040.5 mol).

Now, convert moles to mass:
\[
\text{Mass of H₂SO₄} = 7040.5 \times 98.08 \, \text{g/mol} = 690616 \, \text{g} = 690.62 \, \text{kg}
\]

Since the H₂SO₄ is 92.5% pure, the actual amount of H₂SO₄ needed is:
\[
\text{Mass of H₂SO₄} = \frac{690.62}{0.925} = 746.06 \, \text{kg}
\]

### Final Answer:
- **NaCl required**: 941.51 kg
- **H₂SO₄ required**: 746.06 kg

Do you want more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How would the required quantities change if the NaCl decomposition degree was different?
2. What is the theoretical yield of HCl in this reaction?
3. How would impurities in Na₂SO₄ affect the mass balance calculation?
4. Can we calculate the energy required for this reaction based on enthalpy values?
5. How does temperature impact the decomposition rate of NaCl in this reaction?

**Tip:** Always verify the purity of reactants in a chemical process to ensure accurate mass balance calculations.

Calculate mass balance of hydrochloric-sulfate production for 1 ton sodium sulfate. Table salt, containing 95% NaCl, and vitriol oil, containing 92.5% H₂SO₄, are used in production. The NaCl decomposition degree is 92%.

This article explains how to calculate the mass balance for producing 1 ton of sodium sulfate using NaCl and H₂SO₄, accounting for 95% purity NaCl, 92.5% purity H₂SO₄, and a 92% NaCl decomposition degree. Includes molar mass calculations and stoichiometry.

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