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AnnZ [28]
3 years ago
6

4Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)

Chemistry
1 answer:
Novay_Z [31]3 years ago
5 0

Explanation:

The number acquired by an element after the lose or gain of an electron is called oxidation number.

For example, 4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)

Here, oxidation number of Fe(s) is 0 and Fe in Fe_{2}O_{3} is +3.

Oxidation number of O in O_{2}(g) is 0 as it is present in its elemental state.

The oxidation number of O in Fe_{2}O_{3} is calculated as follows.

2(3) + 3x = 0\\6 + 3x = 0\\x = \frac{-6}{3}\\= -2

Hence,  oxidation number of O in Fe_{2}O_{3} is -2.

  • The loss of electrons by an element or substance is called oxidation. Here, electrons are being lost by Fe(s) as an increase in oxidation state is occurring. So, Fe(s) is oxidized.
  • The gain of electrons by an element or substance is called reduction. Here, electrons are being added to O_{2} as a decrease in its oxidation state is occurring. So, O_{2}  is reduced.
  • An element or compound which is being reduced is called oxidizing agent. Here, O_{2}  is the oxidizing agent.
  • An element or compound which is being oxidized is called reducing agent. Here, Fe(s) is the reducing agent.
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A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

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Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

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