If 10.0 L of butane and oxygen are combined at the same temperature and pressure, which gas will be left over after reaction? Include your calculation/reasoning.
To determine which gas will be left over after the reaction, we first need to write the balanced chemical equation for the reaction between butane (C4H10) and oxygen (O2):
C4H10 + O2 -> CO2 + H2O
From the balanced equation, we can see that 1 mole of butane reacts with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.
Now, let's calculate the number of moles of butane and oxygen in the given 10.0 L:
To calculate the number of moles, we'll use the ideal gas law equation:
PV = nRT
Where:
P = pressure (assumed constant)
V = volume (10.0 L)
n = number of moles (what we want to find)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (assumed constant)
Since the temperature and pressure are constant, we can simplify the equation to:
V = n (at constant temperature and pressure)
For butane:
V (butane) = n (butane)
For oxygen:
V (oxygen) = n (oxygen)
Given that we have 10.0 L of both butane and oxygen, and assuming both gases are at the same temperature and pressure, we can conclude that the number of moles of butane is equal to the number of moles of oxygen in the initial mixture.
Therefore, after the reaction, both gases will be completely consumed. There will be no gas left over.
Note: This assumption holds if the reaction goes to completion and there are no other limiting factors or side reactions occurring.