Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Find the volume of 0.130M sulfuric acid necessary to react completely with 65.9g sodium hydroxide

Sagot :

Eduard22sly

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

Learn more: https://brainly.com/question/7882345

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.