Answer :
To solve this problem, we need to follow a systematic approach based on stoichiometry. Here's the detailed, step-by-step solution:
1. Write the balanced chemical equation:
[tex]\[ 2 \, \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} \][/tex]
This equation shows that 1 mole of sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\)[/tex]) reacts with 2 moles of sodium hydroxide ([tex]\(\text{NaOH}\)[/tex]) to produce 1 mole of sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]) and 2 moles of water ([tex]\(\text{H}_2\text{O}\)[/tex]).
2. Determine the molar masses:
- Molar mass of sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\)[/tex]):
[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1\, \text{g/mol (H)} + 32\, \text{g/mol (S)} + 4 \times 16\, \text{g/mol (O)} = 2 + 32 + 64 = 98\, \text{g/mol} \][/tex]
- Molar mass of sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]):
[tex]\[ \text{Na}_2\text{SO}_4 = 2 \times 23\, \text{g/mol (Na)} + 32\, \text{g/mol (S)} + 4 \times 16\, \text{g/mol (O)} = 46 + 32 + 64 = 142\, \text{g/mol} \][/tex]
3. Calculate the moles of sulfuric acid used:
Given the mass of sulfuric acid is 355 grams:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{355\, \text{g}}{98\, \text{g/mol}} = 3.622 \, \text{moles} \, (\text{to three significant figures}) \][/tex]
4. Determine the moles of sodium sulfate produced:
From the balanced equation, 1 mole of [tex]\(\text{H}_2\text{SO}_4\)[/tex] produces 1 mole of [tex]\(\text{Na}_2\text{SO}_4\)[/tex]. Thus, 3.622 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex] will produce:
[tex]\[ \text{Moles of } \text{Na}_2\text{SO}_4 = 3.622 \, \text{moles} \][/tex]
5. Calculate the mass of sodium sulfate produced:
[tex]\[ \text{Mass of } \text{Na}_2\text{SO}_4 = \text{Moles of } \text{Na}_2\text{SO}_4 \times \text{Molar mass of } \text{Na}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Mass of } \text{Na}_2\text{SO}_4 = 3.622 \, \text{moles} \times 142\, \text{g/mol} = 514\, \text{g} \, (\text{to three significant figures}) \][/tex]
Therefore, the theoretical mass of sodium sulfate produced is [tex]\(\boxed{514}\)[/tex] grams.
1. Write the balanced chemical equation:
[tex]\[ 2 \, \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} \][/tex]
This equation shows that 1 mole of sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\)[/tex]) reacts with 2 moles of sodium hydroxide ([tex]\(\text{NaOH}\)[/tex]) to produce 1 mole of sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]) and 2 moles of water ([tex]\(\text{H}_2\text{O}\)[/tex]).
2. Determine the molar masses:
- Molar mass of sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\)[/tex]):
[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1\, \text{g/mol (H)} + 32\, \text{g/mol (S)} + 4 \times 16\, \text{g/mol (O)} = 2 + 32 + 64 = 98\, \text{g/mol} \][/tex]
- Molar mass of sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]):
[tex]\[ \text{Na}_2\text{SO}_4 = 2 \times 23\, \text{g/mol (Na)} + 32\, \text{g/mol (S)} + 4 \times 16\, \text{g/mol (O)} = 46 + 32 + 64 = 142\, \text{g/mol} \][/tex]
3. Calculate the moles of sulfuric acid used:
Given the mass of sulfuric acid is 355 grams:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{355\, \text{g}}{98\, \text{g/mol}} = 3.622 \, \text{moles} \, (\text{to three significant figures}) \][/tex]
4. Determine the moles of sodium sulfate produced:
From the balanced equation, 1 mole of [tex]\(\text{H}_2\text{SO}_4\)[/tex] produces 1 mole of [tex]\(\text{Na}_2\text{SO}_4\)[/tex]. Thus, 3.622 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex] will produce:
[tex]\[ \text{Moles of } \text{Na}_2\text{SO}_4 = 3.622 \, \text{moles} \][/tex]
5. Calculate the mass of sodium sulfate produced:
[tex]\[ \text{Mass of } \text{Na}_2\text{SO}_4 = \text{Moles of } \text{Na}_2\text{SO}_4 \times \text{Molar mass of } \text{Na}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Mass of } \text{Na}_2\text{SO}_4 = 3.622 \, \text{moles} \times 142\, \text{g/mol} = 514\, \text{g} \, (\text{to three significant figures}) \][/tex]
Therefore, the theoretical mass of sodium sulfate produced is [tex]\(\boxed{514}\)[/tex] grams.