what volume of oxygen gas is needed to completely combust 0.202 L of butane (C4H10) gas?
1. Write the equation and balance it.
2C4H10 + 13O2 ==> 8CO2 + 10H2O
2.You CAN go through the general stoichiometric solution; however, when everthing is in the form of gases, 2 moles C4H10 gas will required 13 moles O2 gas or
0.202 L x 13/2 = ?? L. The conversion is L to L.
To determine the volume of oxygen gas required to completely combust butane (C4H10) gas, we need to balance the chemical equation for the combustion reaction:
C4H10 + O2 → CO2 + H2O
The balanced equation shows that one molecule of butane (C4H10) reacts with eight molecules of oxygen (O2) to produce four molecules of carbon dioxide (CO2) and five molecules of water (H2O).
Since the volume of butane (C4H10) is given in liters (0.202 L), we can directly compare it to the volume of oxygen gas.
Using the stoichiometry of the balanced equation, we can establish a ratio between the volumes of the reactants. The balanced equation tells us that one volume of butane (C4H10) reacts with eight volumes of oxygen (O2).
Therefore, the volume of oxygen gas needed can be calculated as follows:
Volume of oxygen gas = Volume of butane gas × (8/1)
= 0.202 L × 8
= 1.616 L
Therefore, 1.616 liters of oxygen gas are needed to completely combust 0.202 liters of butane (C4H10) gas.
To determine the volume of oxygen gas needed to completely combust 0.202 L of butane gas (C4H10), we first need to know the balanced chemical equation for the combustion of butane.
The balanced equation for the combustion of butane is:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen gas to produce 8 moles of carbon dioxide and 10 moles of water.
To find the volume of oxygen gas needed, we need to convert the volume of butane from liters to moles using the ideal gas law equation:
PV = nRT
Where:
P = Pressure (assumed to be constant)
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature (assumed to be constant)
Since we are assuming constant pressure and temperature, we can rewrite the equation as:
V = n(RT)/P
Now we need to calculate the number of moles of butane gas. To do this, we can use the ideal gas law equation again:
PV = nRT
Rearranging the equation, we have:
n = PV/RT
Assuming that the pressure, temperature, and ideal gas constant are constant, we can calculate the number of moles of butane gas using its known volume (0.202 L).
Now, since the balanced equation shows that 2 moles of butane react with 13 moles of oxygen gas, we can calculate the moles of oxygen gas required using the stoichiometry of the equation.
Finally, we can convert the moles of oxygen gas to volume using the ideal gas law equation:
V = n(RT)/P
By plugging in the appropriate values for pressure, temperature, and the number of moles of oxygen gas, we can calculate the volume of oxygen gas needed to completely combust 0.202 L of butane gas.