What volume of oxygen(in Liters) at 25 degrees C and 1.05 atm is needed to completely combust 2.65L of methane, measured at STP
CH4 + 2O2 ==> CO2 + 2H2O
You may use a shortcut when dealing with gases and use volume as if they were mols directly. Therefore, we will use 2.65 L CH4. Convert to L O2.
2.65 L CH4 x (2 mols O2/1 mols CH4) = 2.65 x 2/1 = ? L O2.
Then convert ?L O2 at STP to L O2 at the conditions of the problem.
To find the volume of oxygen needed to completely combust methane, we first need to balance the chemical equation for the combustion reaction of methane. The balanced reaction equation is:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).
Since we are given the volume of methane measured at STP, we need to convert it to moles. To convert from volume to moles, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
At STP (Standard Temperature and Pressure), the conditions are 0 degrees Celsius (273.15 Kelvin) and 1 atm. However, in this case, the temperature given is 25 degrees Celsius, so we need to convert it to Kelvin by adding 273.15.
Given:
Volume of methane (V) = 2.65 L
Temperature (T) = 25 degrees Celsius = 25 + 273.15 = 298.15 K
Pressure (P) = 1.05 atm
First, calculate the moles of methane using the ideal gas law:
n(CH4) = (P * V) / (R * T)
= (1.05 atm * 2.65 L) / (0.0821 L·atm/(mol·K) * 298.15 K)
Calculate:
n(CH4) = 0.1132 moles
Since the molar ratio between methane and oxygen is 1:2, we need twice as many moles of oxygen as moles of methane.
n(O2) = 2 * n(CH4)
= 2 * 0.1132 moles
= 0.2264 moles
Now, we need to find the volume of oxygen in liters using the ideal gas law. But first, we need to convert the pressure and temperature to the values given in the question.
Temperature (T) = 25 degrees Celsius = 25 + 273.15 = 298.15 K
Pressure (P) = 1.05 atm
Next, apply the ideal gas law equation again:
V = (n * R * T) / P
= (0.2264 moles * 0.0821 L·atm/(mol·K) * 298.15 K) / 1.05 atm
Calculate:
V = 5.66 L
Therefore, the volume of oxygen needed to completely combust 2.65 L of methane measured at STP is approximately 5.66 L at 25 degrees Celsius and 1.05 atm.