In the following reaction, how many liters of oxygen will react with 124.5 liters of propane (C3H8), at STP?
C3H8 + 5O2 = 3CO2 + 4H2O
Since all are gases one may use a shortcut that tells us we can use liters directly as moles. All you need to do is to convert 124.5 L propane to oxygen using the coefficients in the balanced equation.
124.5L x (5 moles propane/1 mol oxygen) = ? L O2.
662.5 liters
622.5 not 662.5
662.5
To determine how many liters of oxygen will react with 124.5 liters of propane (C3H8), at STP, we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation is:
C3H8 + 5O2 = 3CO2 + 4H2O
From the equation, we can see that for every one mole of propane (C3H8), we need 5 moles of oxygen (O2). This means that the mole ratio between propane and oxygen is 1:5.
First, we need to convert the given liters of propane (C3H8) to moles. To do this, we use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume in liters (124.5 L in this case)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (at STP, it is 273 K)
Rearranging the equation to solve for n (number of moles):
n = PV / RT
Plugging in the values:
n = (1 atm * 124.5 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 5.113 moles of propane (C3H8)
Since the mole ratio between propane and oxygen is 1:5, we multiply the number of moles of propane by 5 to find the number of moles of oxygen required:
n_oxygen = 5 * n_propane
n_oxygen ≈ 5 * 5.113 moles
n_oxygen ≈ 25.57 moles of oxygen (O2)
Now, to convert moles back to liters of oxygen at STP, we use the ideal gas law again:
n = PV / RT
Rearranging the equation to solve for V (volume in liters):
V = nRT / P
Plugging in the values:
V = (25.57 moles * 0.0821 L·atm/mol·K * 273 K) / (1 atm)
V ≈ 569.14 liters of oxygen
Therefore, approximately 569.14 liters of oxygen will react with 124.5 liters of propane (C3H8) at STP.