Physics, 16.01.2021 18:30 schapethan

A mixture of 5 cm3 of CH4 and 100 cm3 of air is exploded. Assume air is 80% N2 by volume and 20% O2 by volume. The resulting mixture is cooled. All volumes are measured at room temperature and pressure. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

What is the composition of the resulting gas?

A mixture of 5 cm3 of CH4 and 100 cm3 of air is exploded. Assume air is 80% N2 by volume and 20% O2

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Answer from: garrismunoz

The final composition of all the gases in the mixture has been $\rm 80\;cm^3$ Nitrogen, $\rm 10\;cm^3$ oxygen, $\rm 5\;cm^3$ carbon dioxide, and $\rm 10\;cm^3$ water (steam). Thus, the correct option is A.

The mixture of air has been composed of 20 % oxygen and 80 % nitrogen. The volume of air has been $\rm 100\;cm^3$ . Thus, the volume of oxygen and nitrogen in the air has been:

The volume of oxygen, $V_O_2=20\;\rm cm^3$

The volume of nitrogen, $V_N_2=80\;\rm cm^3$

Since the reaction, the nitrogen has been remained unused, thus the final composition of nitrogen has been $\rm 80 \;cm^3$ .

The balanced equation for the reaction of methane with oxygen has been:

$\rm CH_4\;+\;2\;O_2\;\rightarrow\;CO_2\;+\;2\;H_2O$

According to Avogadro's law, the presence of gases at the same temperature has volume proportional to the moles of the gases.

Thus, from the balanced chemical equation:

$\rm 1\;cm^3\;CH_4=2\;cm^3\;O_2$

Thus,

$\rm 5\;cm^3\;CH_4=10\;cm^3\;O_2$

Since the available oxygen has been $\rm 20\;cm^3$ , the remaining oxygen after the reaction has been $\rm 10\;cm^3$ .

Thus, the final composition of oxygen has been $\rm 10\;cm^3$ .

Methane has been the limiting reactant. Thus, the concentration of carbon dioxide and water formed has been:

$\rm 1\;cm^3\;CH_4=1\;cm^3\;CO_2\\5\;cm^3\;CH_4=5\;cm^3\;CO_2$

The final composition of carbon dioxide has been $\rm 5\;cm^3$ .

$\rm 1\;cm^3\;CH_4=2\;cm^3\;H_2O\\5\;cm^3\;CH_4=10\;cm^3\;H_2O$

The final composition of water (steam) has been $\rm 10\;cm^3$ .

For more information about the composition of gas, refer to the link:

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Answer from: Derrionna

The composition of the resulting gas;

A. 5 cm³ of CO₂, 10 cm³ of O₂, 80 cm³ of N₂, and 10 cm³

Explanation:

The given parameters are;

The volume of CH₄ in the mixture of air and methane = 5 cm³

The volume of air in the mixture of air and methane = 100 cm³

The percentage by volume of nitrogen, N₂ in the air = 80%

The percentage by volume of oxygen, O₂ in the air = 20%

The equation for the reaction of the chemical reaction between the methane and the air in the mixture is given as follows;

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Therefore, we have;

1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O

By Avogadro's law, we have that equal volumes of all gases at the same temperature and pressure contains equal number of molecules

Therefore, we have;

1 cm₃ of CH₄ reacts with 2 cm³ of O₂ to produce 1 cm³ of CO₂ and 2 cm³ of H₂O

From which we have;

5 cm₃ of CH₄ reacts with 10 cm³ of O₂ to produce 5 cm³ of CO₂ and 10 cm³ of H₂O (steam)

The volume of the oxygen used in the reaction = 10 cm³

The volume of oxygen present in the air = 20% × 100 cm³ = 20 cm³

The volume of nitrogen present in the air = 80% × 100 cm³ = 80 cm³

Therefore, the composition of the resulting gas is given as follows;

1) 80 cm³ Nitrogen

2) 10 cm³ Oxygen

3) 5 cm³ CO₂

4) 10 cm³ H₂O

Answer from: Quest

explanation:

Answer from: Quest

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