If 1.00 kg of methane (CH4) is burned completely (in excess O2) and the CO2 produced is collected at 20°C and 1.00 atm, what volume (L) will this CO2 occupy??
To determine the volume of CO2 produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (1.00 atm)
V = volume in liters (to be determined)
n = number of moles of gas (to be determined)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (20°C = 293.15 K)
First, let's calculate the number of moles of CO2 produced. We start with the balanced chemical equation for the combustion of methane.
CH4 + 2O2 -> CO2 + 2H2O
For every mole of CH4, one mole of CO2 is produced. The molar mass of CH4 is 16.04 g/mol, so 1.00 kg is equal to 1000 g. We can convert the mass of CH4 to moles using its molar mass.
moles of CH4 = (mass of CH4) / (molar mass of CH4)
moles of CH4 = 1000 g / 16.04 g/mol
Now that we have the moles of CH4, we know that the same number of moles of CO2 will be produced. Therefore, n (moles of CO2) is equal to the moles of CH4.
Next, we can substitute the values into the ideal gas law equation:
(P)V = (n)(R)(T)
Since we are solving for V, rearranging the equation, we get:
V = (n)(R)(T) / P
Now we can substitute the known values into the equation:
V = (n)(R)(T) / P
V = (moles of CO2)(0.0821 L·atm/(mol·K))(293.15 K) / 1.00 atm
By plugging in the value for moles of CO2, we can calculate the volume (V).