Let's solve the problem of determining the maximum mass of water that could be produced by the reaction between sulfuric acid ([tex]\( \text{H}_2 \text{SO}_4 \)[/tex]) and sodium hydroxide ([tex]\( \text{NaOH} \)[/tex]).
1. 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]
2. Determine the molar masses:
- Molar mass of [tex]\( \text{H}_2 \text{SO}_4 \)[/tex]:
[tex]\[
2(1.01) + 32.07 + 4(16.00) = 98.08 \text{ g/mol}
\][/tex]
- Molar mass of [tex]\( \text{NaOH} \)[/tex]:
[tex]\[
22.99 + 15.999 + 1.01 = 40.00 \text{ g/mol}
\][/tex]
- Molar mass of [tex]\( \text{H}_2 \text{O} \)[/tex]:
[tex]\[
2(1.01) + 16.00 = 18.015 \text{ g/mol}
\][/tex]
3. Calculate the number of moles of each reactant:
- Moles of [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] (given 16.0 g):
[tex]\[
\frac{16.0 \text{ g}}{98.08 \text{ g/mol}} = 0.163 \text{ mol}
\][/tex]
- Moles of [tex]\( \text{NaOH} \)[/tex] (given 6.14 g):
[tex]\[
\frac{6.14 \text{ g}}{40.00 \text{ g/mol}} = 0.1535 \text{ mol}
\][/tex]
4. Identify the limiting reactant:
- According to the stoichiometry, 1 mole of [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] reacts with 2 moles of [tex]\( \text{NaOH} \)[/tex].
- Moles of [tex]\( \text{NaOH} \)[/tex] required for 0.163 mol [tex]\( \text{H}_2 \text{SO}_4 \)[/tex]:
[tex]\[
0.163 \text{ mol} \times 2 = 0.326 \text{ mol}
\][/tex]
- Since 0.1535 mol of [tex]\( \text{NaOH} \)[/tex] is given, it is less than 0.326 mol, making [tex]\( \text{NaOH} \)[/tex] the limiting reactant.
5. Calculate the moles of water produced:
- According to the reaction, 2 moles of [tex]\( \text{NaOH} \)[/tex] produce 2 moles of [tex]\( \text{H}_2 \text{O} \)[/tex].
- Moles of [tex]\( \text{H}_2 \text{O} \)[/tex] produced by 0.1535 mol of [tex]\( \text{NaOH} \)[/tex]:
[tex]\[
\frac{0.1535 \text{ mol}}{2} = 0.07675 \text{ mol}
\][/tex]
6. Calculate the mass of water produced:
- Using the molar mass of [tex]\( \text{H}_2 \text{O} \)[/tex]:
[tex]\[
0.07675 \text{ mol} \times 18.015 \text{ g/mol} = 1.38 \text{ g}
\][/tex]
Thus, the maximum mass of water that could be produced by the chemical reaction is 1.38 g (rounded to 3 significant digits).
