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
