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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.

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  1. 17 October, 03:55
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    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
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