To find the mass of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] formed when 196 grams of 50% pure [tex]\( \text{H}_2\text{SO}_4 \)[/tex] reacts with excess [tex]\( \text{NaOH} \)[/tex], we can follow these steps:
1. Calculate the mass of pure [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[
\text{Mass of pure } \text{H}_2\text{SO}_4 = 196 \, \text{g} \times 0.50 = 98 \, \text{g}
\][/tex]
2. Determine the molar mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[
\text{Molar mass of } \text{H}_2\text{SO}_4 = 98.079 \, \text{g/mol}
\][/tex]
3. Calculate the number of moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[
\text{Moles of } \text{H}_2\text{SO}_4 = \frac{98 \, \text{g}}{98.079 \, \text{g/mol}} \approx 0.999
\][/tex]
4. Write the balanced chemical equation:
[tex]\[
\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\][/tex]
According to this equation, 1 mole of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] will produce 1 mole of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex].
5. Determine the molar mass of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex]:
[tex]\[
\text{Molar mass of } \text{Na}_2\text{SO}_4 = 142.04 \, \text{g/mol}
\][/tex]
6. Calculate the mass of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] formed:
[tex]\[
\text{Mass of } \text{Na}_2\text{SO}_4 = 0.999 \, \text{mol} \times 142.04 \, \text{g/mol} \approx 141.93 \, \text{g}
\][/tex]
Therefore, the mass of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] formed is approximately 141.93 grams.
