To solve the problem of how much mass of 83.4% pure salt cake (Na₂SO₄) can be produced from 250 kg of 94.5% pure salt (NaCl), we will follow these steps:
### Step 1: Calculate the mass of pure NaCl in the given sample
We start with the total mass of the salt sample and its purity percentage.
\[
\text{Mass of pure NaCl} = \text{Total mass} \times \left( \frac{\text{Purity percentage}}{100} \right)
\]
Substituting the values:
\[
\text{Mass of pure NaCl} = 250 \, \text{kg} \times \left( \frac{94.5}{100} \right) = 236.25 \, \text{kg}
\]
### Step 2: Determine the molar mass of NaCl
Next, we calculate the molar mass of NaCl.
\[
\text{Molar mass of NaCl} = \text{Molar mass of Na} + \text{Molar mass of Cl} = 23 \, \text{g/mol} + 35.5 \, \text{g/mol} = 58.5 \, \text{g/mol}
\]
### Step 3: Calculate the number of moles of NaCl
Now, we convert the mass of pure NaCl into moles.
\[
\text{Number of moles of NaCl} = \frac{\text{Mass of pure NaCl}}{\text{Molar mass of NaCl}} = \frac{236.25 \, \text{kg} \times 1000 \, \text{g/kg}}{58.5 \, \text{g/mol}} = 4041.03 \, \text{mol}
\]
### Step 4: Use stoichiometry to find moles of Na₂SO₄ produced
From the balanced chemical equation:
\[
2 \, \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{HCl}
\]
We see that 2 moles of NaCl produce 1 mole of Na₂SO₄. Therefore, the number of moles of Na₂SO₄ produced is:
\[
\text{Moles of Na}_2\text{SO}_4 = \frac{\text{Moles of NaCl}}{2} = \frac{4041.03 \, \text{mol}}{2} = 2020.515 \, \text{mol}
\]
### Step 5: Calculate the mass of Na₂SO₄ produced
Next, we calculate the molar mass of Na₂SO₄.
\[
\text{Molar mass of Na}_2\text{SO}_4 = (23 \times 2) + 32 + (16 \times 4) = 46 + 32 + 64 = 142 \, \text{g/mol}
\]
Now, we can find the total mass of Na₂SO₄ produced.
\[
\text{Mass of Na}_2\text{SO}_4 = \text{Number of moles} \times \text{Molar mass} = 2020.515 \, \text{mol} \times 142 \, \text{g/mol} = 287,000.23 \, \text{g} = 287 \, \text{kg}
\]
### Step 6: Calculate the mass of 83.4% pure Na₂SO₄
Finally, we need to find out how much of this mass is 83.4% pure Na₂SO₄.
\[
\text{Mass of 83.4% pure Na}_2\text{SO}_4 = \frac{\text{Mass of Na}_2\text{SO}_4}{\text{Purity percentage}} = \frac{287 \, \text{kg}}{0.834} \approx 344.00 \, \text{kg}
\]
### Final Answer
The mass of 83.4% pure salt cake (Na₂SO₄) produced is approximately **344 kg**.
