To solve the problem of how many grams of 90% pure sodium sulfate (Na2SO4) can be produced from 250 g of 95% pure sodium chloride (NaCl), we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between sodium chloride and sulfuric acid is as follows:
\[
2 \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCl}
\]
### Step 2: Determine the amount of pure NaCl
Given that we have 250 g of 95% pure NaCl, we need to calculate the mass of pure NaCl:
\[
\text{Mass of pure NaCl} = 250 \, \text{g} \times \frac{95}{100} = 237.5 \, \text{g}
\]
### Step 3: Calculate the moles of NaCl
The molar mass of NaCl is approximately 58.5 g/mol. We can calculate the number of moles of NaCl:
\[
\text{Moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{237.5 \, \text{g}}{58.5 \, \text{g/mol}} \approx 4.05 \, \text{mol}
\]
### Step 4: Use stoichiometry to find moles of Na2SO4 produced
From the balanced equation, 2 moles of NaCl produce 1 mole of Na2SO4. Therefore, the moles of Na2SO4 produced from 4.05 moles of NaCl is:
\[
\text{Moles of Na2SO4} = \frac{4.05 \, \text{mol NaCl}}{2} = 2.025 \, \text{mol Na2SO4}
\]
### Step 5: Calculate the mass of Na2SO4 produced
The molar mass of Na2SO4 is approximately 142 g/mol. Thus, the mass of Na2SO4 produced is:
\[
\text{Mass of Na2SO4} = \text{Moles of Na2SO4} \times \text{Molar mass of Na2SO4} = 2.025 \, \text{mol} \times 142 \, \text{g/mol} \approx 288.25 \, \text{g}
\]
### Step 6: Calculate the mass of 90% pure Na2SO4
Since we need the mass of 90% pure Na2SO4, we calculate it as follows:
\[
\text{Mass of 90% pure Na2SO4} = 288.25 \, \text{g} \times \frac{90}{100} = 259.425 \, \text{g}
\]
### Final Answer
Rounding to one decimal place, the mass of 90% pure sodium sulfate produced is approximately:
\[
\text{Mass of 90% pure Na2SO4} \approx 259.4 \, \text{g}
\]
### Summary
Thus, the final answer is **259.4 g** of 90% pure sodium sulfate can be produced from 250 g of 95% pure sodium chloride.
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