To determine how many moles of water are formed when 15 mL of 0.40 M LiOH (Lithium hydroxide) is added to 25 mL of 0.60 M H₂SO₄ (Sulfuric acid), follow these detailed steps.
1. Convert the volumes from milliliters to liters:
- Volume of LiOH: 15 mL = [tex]\( \frac{15}{1000} \)[/tex] L = 0.015 L
- Volume of H₂SO₄: 25 mL = [tex]\( \frac{25}{1000} \)[/tex] L = 0.025 L
2. Calculate the moles of LiOH and H₂SO₄:
- Moles of LiOH = Concentration (M) × Volume (L)
[tex]\[
\text{Moles of LiOH} = 0.40 \, \text{M} \times 0.015 \, \text{L} = 0.006 \, \text{mol}
\][/tex]
- Moles of H₂SO₄ = Concentration (M) × Volume (L)
[tex]\[
\text{Moles of H₂SO₄} = 0.60 \, \text{M} \times 0.025 \, \text{L} = 0.015 \, \text{mol}
\][/tex]
3. Determine the limiting reagent using stoichiometry:
According to the balanced chemical equation:
[tex]\[
2 \, \text{LiOH} + \text{H₂SO₄} \to \text{Li₂SO₄} + 2 \, \text{H₂O}
\][/tex]
- For H₂SO₄:
1 mole of H₂SO₄ reacts with 2 moles of LiOH.
- For LiOH:
2 moles of LiOH are required to react with 1 mole of H₂SO₄.
4. Identify the limiting reagent:
- Moles of H₂SO₄ divided by its stoichiometric coefficient:
[tex]\[
\frac{0.015 \, \text{moles}}{1} = 0.015
\][/tex]
- Moles of LiOH divided by its stoichiometric coefficient:
[tex]\[
\frac{0.006 \, \text{moles}}{2} = 0.003
\][/tex]
Since 0.003 < 0.015, LiOH is the limiting reagent.
5. Calculate the number of moles of water formed:
Since LiOH is the limiting reagent and the stoichiometric ratio of LiOH to H₂O is 2:2 (or 1:1), for every mole of LiOH, 1 mole of H₂O is formed.
- Moles of H₂O = Moles of limiting reagent (LiOH) used in reaction
[tex]\[
\text{Moles of H₂O} = 0.006 \, \text{mol}
\][/tex]
Thus, the number of moles of water formed is [tex]\(\boxed{0.0060 \, \text{mol}}\)[/tex]. The correct answer is B.
