Final answer:
The mass of NaOH required to react with 25.0 mL of 2.7 M H₂SO₄ is calculated using stoichiometry based on the balanced equation between H₂SO₄ and NaOH. One mole of H₂SO₄ reacts with two moles of NaOH. The resulting calculation gives us a mass of 5.40 g of NaOH.
Step-by-step explanation:
To calculate the mass of NaOH required to react with a given volume and concentration of H₂SO₄, we use the balanced chemical equation:
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide. To solve the problem, we start by calculating the number of moles of H₂SO₄:
- Number of moles H₂SO₄ = Volume(L) × Molarity(M) = 0.025 L × 2.7 M = 0.0675 mol H₂SO₄.
- According to the chemical equation, 1 mol H₂SO₄ reacts with 2 mol NaOH, so the moles of NaOH needed = 2 × 0.0675 mol = 0.135 mol NaOH.
- Finally, calculate the mass of NaOH using its molar mass (Molar mass of NaOH = 40.00 g/mol):
- Mass of NaOH = Number of moles × Molar mass = 0.135 mol × 40.00 g/mol = 5.40 g.
Therefore, the mass of NaOH needed to react with 25.0 mL of 2.7 M H₂SO₄ is 5.40 g.