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To solve this problem, we need to perform a mass balance on the reaction that produces sodium sulfate (\( \text{Na}_2\text{SO}_4 \)) and hydrochloric acid (HCl) from sodium chloride (NaCl) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)).

### Reaction involved:
\[
2 \text{NaCl} + \text{H}_2\text{SO}_4} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCl}
\]

### Information provided:
- 1 ton (1000 kg) of sodium sulfate (\( \text{Na}_2\text{SO}_4 \)) is produced.
- Table salt contains 95% NaCl.
- Vitriol oil contains 92.5% \( \text{H}_2\text{SO}_4 \).
- The decomposition degree of NaCl is 92%.

### Step-by-Step Mass Balance:

#### Step 1: Calculate the molar masses
- Molar mass of NaCl = 58.44 g/mol.
- Molar mass of \( \text{H}_2\text{SO}_4 \) = 98.08 g/mol.
- Molar mass of \( \text{Na}_2\text{SO}_4 \) = 142.04 g/mol.

#### Step 2: Moles of sodium sulfate produced
1 ton of sodium sulfate is produced, which equals 1000 kg or \( 1,000,000 \) g.

\[
\text{Moles of } \text{Na}_2\text{SO}_4 = \frac{1,000,000 \text{ g}}{142.04 \text{ g/mol}} = 7,041.6 \text{ mol}
\]

#### Step 3: Moles of NaCl required
From the balanced equation, 2 moles of NaCl produce 1 mole of sodium sulfate. Therefore, the moles of NaCl required are:

\[
\text{Moles of NaCl} = 2 \times 7,041.6 = 14,083.2 \text{ mol}
\]

#### Step 4: Mass of NaCl required
\[
\text{Mass of NaCl} = 14,083.2 \text{ mol} \times 58.44 \text{ g/mol} = 823,125 \text{ g} = 823.125 \text{ kg}
\]

Since only 92% of the NaCl decomposes, we need to divide by 0.92 to get the total amount of NaCl required:

\[
\text{Actual NaCl required} = \frac{823.125}{0.92} = 894.7 \text{ kg}
\]

#### Step 5: Adjust for purity of NaCl
Since the NaCl is only 95% pure, we adjust for purity:

\[
\text{Mass of salt required} = \frac{894.7}{0.95} = 941.79 \text{ kg}
\]

#### Step 6: Calculate the mass of \( \text{H}_2\text{SO}_4 \) required
From the balanced equation, 1 mole of \( \text{H}_2\text{SO}_4 \) is required for every 2 moles of NaCl. Therefore, the moles of \( \text{H}_2\text{SO}_4 \) required are:

\[
\text{Moles of } \text{H}_2\text{SO}_4 = \frac{14,083.2}{2} = 7,041.6 \text{ mol}
\]

\[
\text{Mass of } \text{H}_2\text{SO}_4 = 7,041.6 \times 98.08 \text{ g/mol} = 690,582 \text{ g} = 690.58 \text{ kg}
\]

Since the sulfuric acid is only 92.5% pure, we need to adjust for purity:

\[
\text{Actual mass of vitriol oil} = \frac{690.58}{0.925} = 746.06 \text{ kg}
\]

### Summary:
- Mass of salt (95% NaCl) required = 941.79 kg.
- Mass of vitriol oil (92.5% \( \text{H}_2\text{SO}_4 \)) required = 746.06 kg.

Let me know if you'd like further details or clarification! 

### Related questions:
1. How does the decomposition degree of NaCl affect the mass balance?
2. How would the mass balance change if the purity of sulfuric acid were different?
3. What is the significance of reaction stoichiometry in mass balance calculations?
4. How would incomplete decomposition of \( \text{H}_2\text{SO}_4 \) influence the mass balance?
5. How do you account for byproducts or impurities in industrial chemical reactions?

### Tip:
In mass balance problems, always ensure to account for the purity and yield of the reactants to avoid underestimating or overestimating the required quantities.

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%.

Calculate the mass balance for the production of sodium sulfate (Na2SO4) from NaCl and H2SO4. The problem involves key concepts like stoichiometry, decomposition degree of NaCl (92%), and purity adjustments for table salt (95% NaCl) and vitriol oil (92.5% H2SO4). Learn how to compute the required quantities for industrial chemical reactions.

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