Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O).
Write a balanced equation for this reaction.
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of oxygen will be required to completely burn 0.700 L of propane gas?
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of carbon dioxide gas will be produced in the reaction?
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of water vapor will be produced in the reaction
The point of this problem is to illustrate that when all are gases one may use a shortcut in which L is used as if L were mols. So using the coefficients in the balanced equation to convert L of one substance to L of any other. I assume you can balance the equation.
The balanced equation for the combustion of propane is as follows:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, we can see that 1 mole of propane requires 5 moles of oxygen to completely burn.
To find the number of liters of oxygen required, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at constant temperature)
V = volume (in liters)
n = number of moles
R = ideal gas constant
T = temperature (in Kelvin)
Since volume measurements occur at the same temperature and pressure, we can assume constant pressure (P) and temperature (T):
PV = nRT
Dividing both sides by RT, we get:
V = n/RT
To find the number of moles of oxygen (O2) needed, we can use the stoichiometric ratio from the balanced equation. Since 1 mole of propane requires 5 moles of oxygen, we have:
n(O2) = 5 * n(C3H8)
Given that 0.700 L of propane gas is burned, we need to convert this volume into moles:
n(C3H8) = 0.700 L * (1 mole/22.4 L) = 0.03125 moles
Now we can calculate the number of moles of oxygen:
n(O2) = 5 * 0.03125 moles = 0.15625 moles
Finally, to find the number of liters of oxygen required:
V(O2) = n(O2) / (R * T)
Let's assume the temperature is 298 K and the ideal gas constant (R) is 0.0821 L.atm/(mol.K):
V(O2) = 0.15625 moles / (0.0821 L.atm/(mol.K) * 298 K) = 0.00702 L
Therefore, approximately 0.00702 liters of oxygen will be required to completely burn 0.700 L of propane gas.
To calculate the number of liters of carbon dioxide (CO2) produced, we use the stoichiometric ratio from the balanced equation. 1 mole of propane produces 3 moles of carbon dioxide, so:
n(CO2) = 3 * n(C3H8) = 3 * 0.03125 moles = 0.09375 moles
To convert this into liters at the same temperature and pressure:
V(CO2) = 0.09375 moles / (0.0821 L.atm/(mol.K) * 298 K) = 0.00418 L
Approximately 0.00418 liters of carbon dioxide gas will be produced in the reaction.
Similarly, to find the number of liters of water vapor (H2O) produced, we use the stoichiometric ratio from the balanced equation. 1 mole of propane produces 4 moles of water vapor, so:
n(H2O) = 4 * n(C3H8) = 4 * 0.03125 moles = 0.125 moles
Using the same calculations as before:
V(H2O) = 0.125 moles / (0.0821 L.atm/(mol.K) * 298 K) = 0.00557 L
Approximately 0.00557 liters of water vapor will be produced in the reaction.
To write a balanced equation for the reaction, you need to ensure that there are an equal number of atoms of each element on both sides of the equation. Here's the balanced equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
Now let's move on to the remaining questions.
To determine the number of liters of oxygen gas required to completely burn 0.700 L of propane gas, we can use the molar ratios from the balanced equation. From the equation, we can see that for every 1 mole of propane gas, we need 5 moles of oxygen. The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol.
First, we need to convert the volume of propane gas from liters to moles. Using the ideal gas equation (PV = nRT), substituting the given temperature and pressure, we can calculate the number of moles of propane gas:
n = PV / RT
= (0.700 L) * (1 atm / 22.4 L) / (0.0821 L•atm/mol•K) * (273 K)
= 0.0261 mol
Now, we use the molar ratio from the balanced equation:
5 moles of oxygen / 1 mole of propane = x moles of oxygen / 0.0261 moles of propane
Simplifying this equation, we find that x = 0.1305 moles of oxygen gas are required.
Finally, we can convert moles of oxygen gas to liters, again using the molar volume at STP:
V = n * 22.4 L/mol
= 0.1305 mol * 22.4 L/mol
= 2.92 L
Therefore, 2.92 liters of oxygen gas will be required to completely burn 0.700 L of propane gas.
To determine the number of liters of carbon dioxide gas produced in the reaction, we again use the molar ratio from the balanced equation. From the equation, we can see that for every 1 mole of propane gas, 3 moles of carbon dioxide gas are produced.
Using the same approach as before, we convert the moles of propane gas to moles of carbon dioxide gas using the molar ratio:
3 moles of carbon dioxide / 1 mole of propane = x moles of carbon dioxide / 0.0261 moles of propane
Simplifying this equation, we find x = 0.0783 moles of carbon dioxide gas produced.
Finally, we can convert moles of carbon dioxide gas to liters again using the molar volume at STP:
V = n * 22.4 L/mol
= 0.0783 mol * 22.4 L/mol
= 1.75 L
Therefore, 1.75 liters of carbon dioxide gas will be produced in the reaction.
To determine the number of liters of water vapor produced in the reaction, we once again use the molar ratio from the balanced equation. From the equation, we can see that for every 1 mole of propane gas, 4 moles of water vapor are produced.
Using the same approach as before, we convert the moles of propane gas to moles of water vapor using the molar ratio:
4 moles of water vapor / 1 mole of propane = x moles of water vapor / 0.0261 moles of propane
Simplifying this equation, we find x = 0.1044 moles of water vapor produced.
Finally, we can convert moles of water vapor to liters again using the molar volume at STP:
V = n * 22.4 L/mol
= 0.1044 mol * 22.4 L/mol
= 2.34 L
Therefore, 2.34 liters of water vapor will be produced in the reaction.