Question

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?

Answer 1

Answer:

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