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? *

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) =

the answer is na (sodium)

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