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

Question

gulaghasi [49]

3 years ago

7

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? *

Chemistry

1 answer:

KiRa [710] · Answer 1 · 2022-05-04T21:42:39+03:00

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:

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

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

<em>Moles H₂SO₄:</em>

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

<em>Mass H₂SO₄ -Molar mass: 98g/mol-</em>

0.4155 moles H₂SO₄ * (98g / mol) =

<h3>40.72g of sulfuric acid are needed</h3>