How many grams of Sulfuric Acid are needed to produce 57.18 g of Lead (IV) Sulfate when being neutralized by a sufficient amount of Lead (IV

Question

How many grams of Sulfuric Acid are needed to produce 57.18 g of Lead (IV) Sulfate when being neutralized by a sufficient amount of Lead (IV) Hydroxide? *

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Kim Chi 6 months 2021-08-05T22:14:49+00:00 1 Answers 8 views 0

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    2021-08-05T22:16:17+00:00

    Answer:

    40.72g of sulfuric acid are needed

    Explanation:

    When sulfuric acid, H₂SO₄, is neutralized by lead (IV) hydroxide, Pb(OH)₄, Lead (IV) sulfate, Pb(SO₄)₂ and water as follows:

    2 H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O

    To solve this question we must find the moles of 57.18g of Pb(SO₄)₂. As 2 moles of H₂SO₄ produce 1mol Pb(SO₄)₂ we can find the moles of H₂SO₄ and its mass as follows:

    Moles Pb(SO₄)₂ -Molar mass: 275.23 g/mol-

    57.18g * (1mol / 275.23g) = 0.2078 moles Pb(SO₄)₂

    Moles H₂SO₄:

    0.2078 moles Pb(SO₄)₂ * (2mol H₂SO₄ / 1mol Pb(SO₄)₂) = 0.4155 moles H₂SO₄

    Mass H₂SO₄ -Molar mass: 98g/mol-

    0.4155 moles H₂SO₄ * (98g / mol) =

    40.72g of sulfuric acid are needed

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