how many liters of oxygen will react with 124.5 liters of propane
To do what? Combust?
C3H8+ 5O2 >> 3CO2 + 4H20
Looks like 1/5 * 124.5 liters
To determine how many liters of oxygen will react with 124.5 liters of propane, we need to know the balanced chemical equation for the combustion reaction between propane (C3H8) and oxygen (O2). The balanced equation is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that it takes 5 moles of oxygen (O2) to react with 1 mole of propane (C3H8).
Step 1: Convert liters of propane (C3H8) to moles
To convert from liters to moles, we need to know the molar volume at a given temperature and pressure. Assuming standard temperature and pressure conditions (STP), the molar volume is 22.4 liters/mol.
124.5 liters of propane (C3H8) / 22.4 liters/mol = 5.56 moles of propane (C3H8)
Step 2: Determine the molar ratio between oxygen and propane
From the balanced equation, we know that the ratio is 5 moles of oxygen (O2) to 1 mole of propane (C3H8).
5.56 moles of propane (C3H8) x (5 moles of oxygen (O2) / 1 mole of propane (C3H8)) = 27.8 moles of oxygen (O2)
Step 3: Convert moles of oxygen to liters
Again, assuming STP, we can convert moles of oxygen to liters using the molar volume.
27.8 moles of oxygen (O2) x 22.4 liters/mol = 622.7 liters of oxygen (O2)
Therefore, 124.5 liters of propane (C3H8) will react with 622.7 liters of oxygen (O2) in the combustion reaction.