12.0cm³ of methane and 48cm³ of to oxygen were exploded together.The final volume of gases was measured at room temperature and pressure.Calculate composition of the resulting gaseous mixture
To calculate the composition of the resulting gaseous mixture, we need to determine the ratio of methane to oxygen in the initial gases.
According to the balanced chemical equation for the combustion of methane:
CH₄ + 2O₂ → CO₂ + 2H₂O
From the equation, we can see that one molecule of methane requires 2 molecules of oxygen to react completely and form 1 molecule of carbon dioxide and 2 molecules of water.
Let's convert the initial volumes of methane and oxygen to moles using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Using the given values:
For methane:
V₁ = 12.0 cm³
P₁ = room pressure
T₁ = room temperature
For oxygen:
V₂ = 48.0 cm³
P₂ = room pressure
T₂ = room temperature
We can cancel out the pressure and temperature from the equation, as they are the same for both gases:
V₁/n₁ = V₂/n₂
Solving for n₁:
n₁ = (V₁/V₂) * n₂
Let's assume the final volume of the gaseous mixture is V₃ cm³. From the balanced equation, we know that 1 molecule of methane reacts to form 1 molecule of carbon dioxide and 2 molecules of water. Hence, the volume of carbon dioxide produced will be equal to the initial volume of methane used (12.0 cm³).
So, V₃ = 12.0 cm³ + V₂ cm³
Now, let's substitute the values to calculate the composition of the resulting gaseous mixture.
n₁ = (12.0 cm³ / (12.0 cm³ + 48.0 cm³)) * n₂
= (12.0 / 60.0) * n₂
= 0.2 * n₂
Since the volume of methane used is equal to the volume of carbon dioxide produced, the volume of oxygen used is equal to twice the volume of carbon dioxide produced.
n₂ = V₂ / (2 * V₃)
= 48.0 cm³ / (2 * (12.0 cm³ + 48.0 cm³))
= 48.0 / (2 * 60.0)
= 0.4
n₁ = 0.2 * n₂
= 0.2 * 0.4
= 0.08
Therefore, the composition of the resulting gaseous mixture is approximately 8% methane and 92% oxygen.