If 10L each of propane and oxygen are combined at the same temperature and pressure, which gas will be left over after reaction? What volume of that gas will remain? The answer is 8 L of Propane
Tried for a while but I don't understand how answer books answer is 8L of Propane.
You may use a shortcut to do these stoichiometry problems when only gases are involved.
C3H8 + 5O2 ==> 3CO2 + 4H2O
I do these in two parts.
1. How much CO2 is produced if I have 10L C3H8 and all the oxygen I need? That is
10L propane x (3 moles CO2/1 L C3H8) = 10 x 3/1 = 30 L CO2
2. How much CO2 is produced if I have 10 L O2 and all the propane I need? That is
10 L O2 x (3 mols CO2/5 mols O2) = 10 x 3/5 = 6 L CO2.
Both answers can't be correct; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, O2 is the limiting reagent (and we can produce 6L CO2 although that's isn't the question). Now that we know the limiting reagent, we can calculate how much of the other one is used.
10 L O2 x (1 mol C3H8/5 mol O2) = 10 x 1/5 = 2 L propane used.
We started with 10, used 2, we must have 8 left over. Book is right.
To determine the volume of gas that will remain after the reaction between propane and oxygen, we need to consider the stoichiometry of the reaction.
The balanced chemical equation for the combustion of propane can be written as:
C3H8 + 5O2 → 3CO2 + 4H2O
From the balanced equation, we can see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).
Given that we have 10 L of propane and 10 L of oxygen, we need to determine which of the two reactants is the limiting reactant, i.e., the reactant that will be completely consumed first.
First, let's convert the volumes to moles using the ideal gas law equation:
n = PV / RT
where:
n = number of moles
P = pressure (assumed to be constant)
V = volume
R = ideal gas constant
T = temperature (assumed to be constant)
Given that the temperature and pressure are the same for both gases, we can simplify the equation to:
n = V / R
For propane: n(C3H8) = 10 L / R
For oxygen: n(O2) = 10 L / R
Since both gases have the same volume, the ratio of moles of propane to moles of oxygen is 1:1.
Therefore, the ratio of moles of propane to moles of oxygen in the reaction is also 1:1.
This means that both propane and oxygen will react completely, and no gas will be left over. Thus, the answer provided in the answer book (8 L of propane remaining) is incorrect.
To understand why the answer is 8 L of propane, let's walk through the steps to solve this problem.
1. Start by writing a balanced chemical equation for the reaction between propane (C3H8) and oxygen (O2):
C3H8 + 5O2 ⟶ 3CO2 + 4H2O
2. Calculate the number of moles of propane and oxygen given the volume of gas provided. We can use the ideal gas law equation to do this:
PV = nRT
Where:
P = pressure (in this case, assume it is constant)
V = volume of gas
n = number of moles
R = ideal gas constant
T = temperature (in this case, assume it is constant)
Assuming temperature and pressure are constant, we can simplify the ideal gas law equation to:
V = n
So, for 10 L of propane, we have 10 moles of propane (C3H8), and for 10 L of oxygen, we have the same number of moles, which is 10 moles of oxygen (O2).
3. Examine the balanced chemical equation to determine the stoichiometry of the reaction. The ratio states that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.
4. Determine which reactant is the limiting reagent. The limiting reagent is the reactant that is used up first and determines the amount of product formed. To do this, compare the moles of propane and oxygen using the stoichiometry of the reaction.
Given that we have 10 moles of propane and 10 moles of oxygen, we can see that both reactants are in the necessary ratio to fully react. This means that neither propane nor oxygen is in excess, and both will be completely consumed in the reaction.
5. To find the volume of the gas that remains, we need to determine the number of moles of the product formed. From the balanced equation, we know that for every 3 moles of carbon dioxide produced, 1 mole of propane is consumed. Therefore, we can calculate the moles of carbon dioxide formed:
moles of carbon dioxide = 10 moles of propane × (3 moles of carbon dioxide / 1 mole of propane)
= 30 moles of carbon dioxide
6. Finally, using the ideal gas law equation, we can determine the volume of carbon dioxide formed, which is equal to the volume of the remaining gas:
V = n
The volume of carbon dioxide formed is 30 L. Since both the reactants are fully consumed in the reaction, the volume of propane remaining is equal to the initial volume of 10 L minus the volume of carbon dioxide formed:
Volume of propane remaining = 10 L - 30 L
= -20 L
It is important to note that the volume cannot be negative, so the answer given in the answer book of 8 L of propane remaining is incorrect. Given the information provided, there should be no propane gas remaining, as it is completely consumed during the reaction.