what volume of oxygen gas is needed to completely combst 0.202L of butane (c4h10)gas?
Write a balanced equation. Here's the equation. Balance the equation yourself.
C4H10 + O2 -> CO2 + H2O
At STP, 1 mol = 22.4L. Use this to convert L to mol butane.
Use stoichiometry to solve this problem. Steps 1-3.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of oxygen gas needed to completely combust butane (C4H10), you need to balance the chemical equation representing the combustion reaction of butane.
The balanced equation for the combustion of butane is as follows:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the balanced equation, you can see that it requires 13 moles of oxygen gas (O2) to react with 2 moles of butane (C4H10).
Now, let's convert the given volume of butane gas (0.202 L) to moles using the ideal gas law, assuming standard temperature and pressure (STP):
PV = nRT
Using the ideal gas law, we can calculate the number of moles of butane:
P (pressure) = 1 atm (at standard pressure)
V (volume) = 0.202 L
R (ideal gas constant) = 0.0821 L·atm/(mol·K)
T (temperature) = 273.15 K (at standard temperature)
n = PV / RT
n = (1 atm * 0.202 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n ≈ 0.0091 mol
Now that we know there are 0.0091 moles of butane gas, we can use the stoichiometry of the balanced equation to find the number of moles of oxygen gas needed.
According to the equation, 2 moles of butane require 13 moles of oxygen gas. Therefore, for 0.0091 moles of butane:
0.0091 mol butane * (13 mol oxygen gas / 2 mol butane) ≈ 0.0593 moles of oxygen gas
Finally, to convert moles of oxygen gas to volume at STP, we can use the ideal gas law one more time:
V = nRT / P
V = (0.0593 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / (1 atm)
V ≈ 1.383 L
Therefore, approximately 1.383 liters of oxygen gas are needed to completely combust 0.202 liters of butane gas.