Question

Using the balanced reaction below, what volume of a 3.00 M solution of sulfuric acid is needed to react completely with 45.0 mL of a 1.50 M solution of NaOH?

H₂SO₄ + 2NaOH --> Na₂SO4 + 2H₂O

Answer 1

To completely react with 45.0 mL of a 1.50 M NaOH solution, 11.25 mL of a 3.00 M H₂SO₄ solution is required based on stoichiometric calculations using the balanced chemical equation.

Step-by-step explanation:

To determine the volume of a 3.00 M solution of sulfuric acid (H₂SO₄) needed to completely react with 45.0 mL of a 1.50 M solution of sodium hydroxide (NaOH), we need to use stoichiometry based on the balanced equation:

H₂SO₄ (aq) + 2NaOH(aq) → Na₂SO₄ (aq) + 2H₂O(l)

First, let's calculate the number of moles of NaOH:

moles NaOH = volume (L) × molarity (M) = 0.045 L × 1.50 M = 0.0675 mol

According to the reaction, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, the number of moles of H₂SO₄ needed is half of the moles of NaOH:

moles H₂SO₄ = 0.0675 mol ÷ 2 = 0.03375 mol

Now, let's find the volume of H₂SO₄ needed:

volume H₂SO₄ = moles / molarity = 0.03375 mol / 3.00 M = 0.01125 L

Convert liters to milliliters:

volume H₂SO₄ = 0.01125 L × 1000 mL/L = 11.25 mL

Therefore, you would need 11.25 mL of the 3.00 M H₂SO₄ solution to completely react with 45.0 mL of the 1.50 M NaOH solution.