Given the balanced neutralization equation from part B, how many moles of potassium hydroxide (KOH) are required to neutralize 4.5 mol of sulfuric acid (H2SO4) ? Assume that the sulfuric acid completely dissociates in water.

Question

Chemistry

Charlee Middleton

17 October, 02:59

Given the balanced neutralization equation from part B, how many moles of potassium hydroxide (KOH) are required to neutralize 4.5 mol of sulfuric acid (H2SO4) ? Assume that the sulfuric acid completely dissociates in water.

+5

Answers (1)

Know the Answer?

Juliette Brandt · Answer 1

9 moles of KOH

Explanation:

Let's begin with the neutralization reaction.

Neutralization reaction always gives as product, water and a salt

2KOH + H₂SO₄ → 2H₂O + K₂SO₄

Ratio in the reactant side is 1:2. We can determine this rule of three:

1 mol of sulfuric acid can react with 2 moles of potassium hydroxide

Therefore, 4.5 moles sulfuric acod wil react with (4.5. 2) / 1 = 9 moles of KOH