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Eduard22sly

The volume of sulphuric acid, H₂SO₄ needed for the complete reaction with 72.2 g of sodium hydroxide, NaOH is 5.31 L

We'll begin by calculating the number of mole in 72.2 g of NaOH. This can be obtained as follow:

Mass of NaOH = 72.2 g

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 72.2 / 40

Mole of NaOH = 1.805 mole

  • Next, we shall determine the number of mole of H₂SO₄ needed to react with 1.805 mole of NaOH.

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

From the balanced equation above,

2 moles of NaOH reacted with 1 mole of H₂SO₄.

Therefore,

1.805 mole of NaOH will react with = 1.805 / 2 = 0.9025 mole of H₂SO₄.

  • Finally, we shall determine the volume of H₂SO₄.

Mole of H₂SO₄ = 0.9025 mole

Molarity of H₂SO₄ = 0.170 M

Volume of H₂SO₄ =?

Volume = mole / Molarity

Volume of H₂SO₄ = 0.9025 / 0.170

Volume of H₂SO₄ = 5.31 L

Thus, the volume of H₂SO₄ needed for the reaction is 5.31 L

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