The volume of the 0.130 M sulfuric acid, H₂SO₄ required to react completely with 65.9 g sodium hydroxide, NaOH is 6.34 L
- We'll begin by calculating the number of mole of in 65.9 g sodium hydroxide, NaOH. This can be obtained as follow:
Mass of NaOH = 65.9 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 65.9 / 40
Mole of NaOH = 1.6475 mole
- Next, we shall determine the number of mole of H₂SO₄ needed to react with 1.6475 mole of NaOH. This can be obtained as follow:
2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O
From the balanced equation above,
2 moles of NaOH reacted with 1 mole of H₂SO₄.
Therefore,
1.6475 mole of NaOH will react with = [tex]\frac{1.6475}{2} \\\\[/tex] = 0.82375 mole of H₂SO₄.
- Finally, we shall determine the volume of 0.130 M sulfuric acid, H₂SO₄ required for the reaction.
Molarity of H₂SO₄ = 0.130 M
Mole of H₂SO₄ = 0.82375 mole
Volume of H₂SO₄ =?
Volume = mole / Molarity
Volume of H₂SO₄ = 0.82375 / 0.130
Volume of H₂SO₄ = 6.34 L
Therefore, the volume of the 0.130 M sulfuric acid, H₂SO₄ required for the reaction is 6.34 L
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