To determine the theoretical mass of sodium sulfate ([tex]\(Na_2SO_4\)[/tex]) produced from the reaction of sulfuric acid ([tex]\(H_2SO_4\)[/tex]) with sodium hydroxide ([tex]\(NaOH\)[/tex]), follow these steps:
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]
From this balanced equation, we see that 1 mole of [tex]\(H_2SO_4\)[/tex] reacts with 2 moles of [tex]\(NaOH\)[/tex] to produce 1 mole of [tex]\(Na_2SO_4\)[/tex].
2. Determine the Molar Masses:
- Molar mass of [tex]\(H_2SO_4\)[/tex]:
[tex]\[
2 \times 1.008 + 32.06 + 4 \times 16.00 = 98.076 \, \text{g/mol}
\][/tex]
- Molar mass of [tex]\(Na_2SO_4\)[/tex]:
[tex]\[
2 \times 22.99 + 32.06 + 4 \times 16.00 = 142.04 \, \text{g/mol}
\][/tex]
3. Calculate the Moles of [tex]\(H_2SO_4\)[/tex] Used:
Given the weight of sulfuric acid is 355 grams:
[tex]\[
\text{moles of } H_2SO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{355 \, \text{g}}{98.076 \, \text{g/mol}} = 3.619641910355235 \, \text{moles}
\][/tex]
4. Determine the Moles of [tex]\(Na_2SO_4\)[/tex] Produced:
According to the balanced equation, 1 mole of [tex]\(H_2SO_4\)[/tex] produces 1 mole of [tex]\(Na_2SO_4\)[/tex]. Therefore, the moles of [tex]\(Na_2SO_4\)[/tex] produced is the same as the moles of [tex]\(H_2SO_4\)[/tex] used:
[tex]\[
\text{moles of } Na_2SO_4 = 3.619641910355235
\][/tex]
5. Calculate the Mass of [tex]\(Na_2SO_4\)[/tex] Produced:
[tex]\[
\text{mass of } Na_2SO_4 = \text{moles} \times \text{molar mass} = 3.619641910355235 \times 142.04 \, \text{g/mol} = 514.1339369468575 \, \text{grams}
\][/tex]
Rounding the answer to three significant figures, the theoretical mass of sodium sulfate ([tex]\(Na_2SO_4\)[/tex]) produced is:
[tex]\[
\boxed{514} \, \text{grams}
\][/tex]
