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PSYCHO15rus [73]

3 years ago

12

A gas mixture at room temperature contains 10.0 mol CO and 12.5 mol O2. (a) Compute the mole fraction of CO in the mixture. (b)

The mixture is then heated, and the CO starts to react with the O2 to give CO2: CO(g) " 1 2 O2(g) 8n CO2(g)

1 answer:

zlopas [31]3 years ago

7 0

Answer:

a. the mole fraction of CO in the mixture of CO and O2.

mole fraction = moles of CO/ Total moles of the mixture

Mole fraction of CO = 10/(10+12.5)=0.444

b. Reaction - CO(g)+½O2(g)→CO2(g)

Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2

So given,

At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.

3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2

This means that unused mols are : 7mols of CO and 11mols of O2

Total product mixture = 3 + 7 + 11 = 21mols

mole fraction of CO = 7/21 = 0.33

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Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

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The amount of oxygen is:

$E = 2 * O$

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So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

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