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IRISSAK [1]
1 year ago
14

You have a mixture of the gases Cl₂ and O₂ along with some N₂ in a container at STP. The moles of Cl₂ and O₂ in the mixture are

equivalent and the density of the mixture is 2.063 g/L. Assuming ideal behavior, what is the mole fraction of N₂ in the mixture?
Chemistry
1 answer:
Harman [31]1 year ago
8 0

The mole fraction of N₂ in the mixture, given the data is 0.225

<h3>How to determine the mass of the mixture</h3>
  • Density of mixture = 2.063 g/L
  • Volume (at STP) = 22.4 L
  • Mass of mixture =?

Density = mass / volume

Cross multiply

Mass = Density × volume

Mass of mixture = 2.063 × 22.4

Mass of mixture = 46.2112 g

<h3>How to determine the mole fraction of N₂</h3>

We'll begin by calculating the mole of N₂ in the mixture. This can be obtained as follow:

  • Mass of mixture = 46.2112 g
  • Let the mole of Cl₂ = y
  • Let the mole of O₂ = Mole of Cl₂ = y
  • Total mole = 1 mole
  • Let the mole of N₂ = x =?

Total mole = Mole of N₂ + Mole of Cl₂ + Mole of O₂

1 = x + y + y

1 = x + 2y

2y = 1 - x

y = (1 - x) / 2

Mass of mixture = mass of N₂ + mass of Cl₂ + mass of O₂

Recall

mass = mole × molar mass

Thus,

Mass of mixture = (mole of N₂ × molar mass) + (mole of Cl₂ × molar mass) + (mole of O₂ × molar mass)

46.2112 = 28x + 71y + 32y

But

y = (1 - x) / 2

46.2112 = 28x + 71(1 - x) / 2 + 32(1 - x) / 2

Multiply through by 2

92.4224 = 56x + 71(1 - x) + 32(1 - x)

92.4224 = 56x + 71 - 71x + 32 - 32x

Collect like terms

92.4224 - 71 - 32 = 56x - 71x - 32x

-10.5776 = -47x

Divide both sides by -47

x = -10.5776 / -47

x = 0.225 mole

Finally, we can determine the mole fraction of N₂. This can be obtained as follow:

  • Mole of N₂ = x = 0.225 mole
  • Total mole = 1 mole
  • Mole fraction of N₂ = ?

Mole fraction = mole / total mole

Mole fraction of N₂ = 0.225 / 1

Mole fraction of N₂ = 0.225

Learn more about mole fraction:

brainly.com/question/2769009

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<em>Structure 2</em>

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Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

  1. Split the reaction into an oxidation and reduction half.
  2. By inspecting, balance the half equations with respect to the charges and atoms.
  3. In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
  4. Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
  5. Multiply both equations with appropriate factors to balance the electrons in the two half equations.
  6. Add up the balanced half equations and cancel out any specie that occur on both sides.
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Solution

                            Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

                      Zn → Zn²⁺                          Oxidation half

                      MnO₄⁻ → Mn²⁺                  Reduction half

Balancing of atoms(in acidic medium)

                     Zn → Zn²⁺

                    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

                   Zn → Zn²⁺ + 2e⁻

                    MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

         Multiply the oxidation half by 5 and reduction half by 2:

                          5Zn → 5Zn²⁺ + 10e⁻

                        2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

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              5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

         5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

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