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?

The final composition of all the gases in the mixture has been Nitrogen,  oxygen,  carbon dioxide, and 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 . Thus, the volume of oxygen and nitrogen in the air has been:

The volume of oxygen,

The volume of nitrogen,

Since the reaction, the nitrogen has been remained unused, thus the final composition of nitrogen has been .

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

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:

Thus,

Since the available oxygen has been , the remaining oxygen after the reaction has been .

Thus, the final composition of oxygen has been .

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

The final composition of carbon dioxide has been .

The final composition of water (steam) has been .

The final composition of all the gases in the mixture has been Nitrogen,  oxygen,  carbon dioxide, and water (steam). Thus, the correct option is A.

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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

this a false

explanation:

the answer is b i think