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
C3H8 + 5O2 ==> 3CO2 + 4H2O
When using gases at the same T and P, one can take a shortcut and use liters as if they were mols. For the problem we simply convert L of one thing into L of the other.
0.7L C3H8 x (5 mols O2/1 mol C3H8) = ? L O2 required.
The other parts are done the same way.
1) Balanced equation
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
The balanced equation for the combustion of propane gas is:
C3H8 + 5O2 -> 3CO2 + 4H2O
According to the equation, for every molecule of propane, 5 molecules of oxygen gas are required to burn it completely.
To find out how many liters of oxygen are required to burn 0.700 L of propane gas, we need to use the ratio of their coefficients in the balanced equation.
From the balanced equation, we know that the ratio of propane to oxygen is 1:5. Given that 0.700 L of propane is used, we can calculate the volume of oxygen required as follows:
0.700 L propane * (5 L oxygen / 1 L propane) = 3.5 L oxygen
Therefore, 3.5 liters of oxygen will be required to completely burn 0.700 L of propane gas.
Similarly, using the balanced equation, we can determine the volume of carbon dioxide and water vapor produced in the reaction.
For carbon dioxide:
0.700 L propane * (3 L CO2 / 1 L propane) = 2.1 L CO2
Therefore, 2.1 liters of carbon dioxide gas will be produced in the reaction.
For water vapor:
0.700 L propane * (4 L H2O / 1 L propane) = 2.8 L H2O
Therefore, 2.8 liters of water vapor will be produced in the reaction.
To write a balanced equation for the combustion of propane (C3H8), you need to ensure that the number of atoms on each side of the equation is equal.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
Now, let's calculate the number of liters of oxygen gas required to burn 0.700 L of propane gas.
First, we need to convert the volumes of gases to the number of moles using the ideal gas law equation: PV = nRT.
Given that both the volume measurements occur at the same temperature and pressure, we can ignore the temperature and pressure variables for this calculation.
Using the equation PV = nRT, where P = pressure, V = volume, n = moles, R = ideal gas constant, and T = temperature, we can write:
PV(O2) = n(O2)RT and PV(C3H8) = n(C3H8)RT
Since the temperature, pressure, and R are the same for both gases, the ratios of P and V to n will also be the same:
PV(O2) / n(O2) = PV(C3H8) / n(C3H8)
We know that 1 mol of any gas occupies 22.4 L at standard temperature and pressure (STP). Therefore, we can write:
PV(O2) / n(O2) = 22.4 L/mol
Now, let's calculate the number of moles of propane gas (C3H8) using the given volume.
n(C3H8) = 0.700 L / 22.4 L/mol
n(C3H8) ≈ 0.03125 mol
Since the balanced equation states that the ratio between propane (C3H8) and oxygen (O2) is 1:5, we know that:
n(O2) = 5 * n(C3H8)
n(O2) = 5 * 0.03125 mol
n(O2) ≈ 0.15625 mol
Finally, we can calculate the volume of oxygen gas (O2) needed to burn 0.700 L of propane gas (C3H8) using the following equation:
V(O2) = n(O2) * 22.4 L/mol
V(O2) ≈ 0.15625 mol * 22.4 L/mol
V(O2) ≈ 3.5 L
Therefore, approximately 3.5 liters of oxygen gas (O2) will be required to completely burn 0.700 L of propane gas (C3H8).
To determine the volume of carbon dioxide (CO2) and water vapor (H2O) produced in the reaction, we can use the stoichiometric coefficients from the balanced equation.
1 mol of propane (C3H8) produces 3 mol of carbon dioxide (CO2) and 4 mol of water vapor (H2O).
Given that we have 0.03125 mol of propane (C3H8) in the reaction:
Volume of CO2 = 3 * 0.03125 mol * 22.4 L/mol
Volume of CO2 ≈ 2.1 L
Volume of H2O = 4 * 0.03125 mol * 22.4 L/mol
Volume of H2O ≈ 2.8 L
Therefore, approximately 2.1 liters of carbon dioxide gas (CO2) and 2.8 liters of water vapor (H2O) will be produced in the reaction.