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koban [17]
2 years ago
12

How many grams of O2 are reacted when 3.5 mols of P203 are formed?

Chemistry
1 answer:
Neporo4naja [7]2 years ago
7 0
1.45 moles of P2O3 contains 1.45 moles of O3 which means 4.35 moles of O:

1.45 × 3 = 4.35 moles of O

Then you can calculate moles of O2 by dividing it by 2:

4.35 moles ÷ 2 = 2.175 moles of O2 needed
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Consider a solution that is 1.3×10−2 M in Ba2+ and 2.0×10−2 M in Ca2+. Ksp(BaSO4)=1.07×10−10 Ksp(CaSO4)=7.10×10−5 If sodium sulf
skad [1K]

Answer:

Explanation:

Ksp(BaSO4)=1.07×10−10

BaSO₄ → Ba²⁺ + SO₄²⁻

1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)

but Ba²⁺ = 1.3×10⁻² M

1.07×10⁻¹⁰  = 1.3×10⁻² M × ( SO₄²⁻)

( SO₄²⁻)  = 1.07×10⁻¹⁰  / 1.3×10⁻² = 0.823 × 10⁻⁸ M

while Ksp(CaSO4)=7.10×10−5

CaSO₄ → Ca²⁺ + SO₄²⁻

7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)

( SO₄²⁻)  = 7.10×10⁻⁵  /  2.0×10⁻² = 3.55 × 10⁻³ M

comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄

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