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Andreas93 [3]
3 years ago
9

When electricity flows through water, the water molecules split into molecules of hydrogen and oxygen gas. How many oxygen (O2)

molecules are produced from the splitting of two molecules of water (H2O) if the balanced equation for the reaction is as follows?
2H2O → 2H2 + O2
Chemistry
2 answers:
ikadub [295]3 years ago
5 0
That would be one oxygen molecule.
larisa [96]3 years ago
5 0

Answer: The number of oxygen molecules produced are 1.

<u>Explanation:</u>

Water is formed by the combination of hydrogen and oxygen atoms.

Water splits into 2 types of molecules when electricity is applied.

The balanced chemical equation for the splitting of water into its molecules follows:

2H_2O\rightarrow 2H_2+O_2

By Stoichiometry of the reaction:

2 moles of water splits to form 2 moles of hydrogen molecule and 1 mole of oxygen molecule.

Or,

2 molecules of water splits to form 2 molecules of hydrogen and 1 molecule of oxygen.

Hence, the number of oxygen molecules produced are 1.

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Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of
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Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n .  R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

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