How many litters of Oxygen will be required to burn 3.8 liters of propane, C3H8?
liters *
C2H8 + 5O2 ==> 2CO2 + 4H2O
When working with gases we can use a shortcut in which we use L as if L = mols. Then we make the conversion form mols propane to mols O2 as usual.
?L O2 = 3.8 L C3H8 x (5 mols O2/1 mol C3H8) = 3.8 x 5 = ?
To find out how many liters of oxygen are required to burn 3.8 liters of propane (C3H8), we need to use the balanced chemical equation for the combustion of propane.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the equation, we can see that for every 1 mole of propane burned, we need 5 moles of oxygen. To calculate the number of moles of propane, we need to use the molar mass of propane.
The molar mass of propane (C3H8) is calculated by adding up the atomic masses of carbon (C) and hydrogen (H):
3*(Atomic mass of C) + 8*(Atomic mass of H)
= 3*(12.01 g/mol) + 8*(1.01 g/mol)
= 36.03 g/mol + 8.08 g/mol
= 44.11 g/mol
To convert the volume of propane from liters to moles, we need to use the ideal gas law. The ideal gas law equation is:
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.
Assuming standard temperature and pressure (STP), which is 273.15 K and 1 atm, the ideal gas constant (R) is 0.0821 L·atm/mol·K.
Using the ideal gas law equation, we can calculate the number of moles of propane:
n = PV / RT
Assuming the pressure and temperature are constant, we have:
n = (3.8 L) / (0.0821 L·atm/mol·K * 273.15 K)
n = 0.1573 mol
Now that we know the number of moles of propane, we can determine the number of moles of oxygen required:
Number of moles of Oxygen = 5 * (Number of moles of Propane)
Number of moles of Oxygen = 5 * 0.1573 mol
Number of moles of Oxygen = 0.7865 mol
Finally, we can convert the number of moles of oxygen back to liters by multiplying it by the molar volume of an ideal gas at STP, which is approximately 22.4 L/mol:
Number of liters of Oxygen = 0.7865 mol * 22.4 L/mol
Number of liters of Oxygen = 17.61 L
Therefore, approximately 17.61 liters of oxygen will be required to burn 3.8 liters of propane (C3H8).