2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
In the equation, if 14 L of ethane (C2H6) and 14 L of oxygen (O2) combined and burned to completion, which gas will be leftover after the reaction, and what is the volume of that gas remaining?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
In all gas systems volume may be used as mols and that cuts out a step or two.
14 L ethane x (4 mols CO2/2 mol C2H6) = about 28 L CO2 possibly formed
14 L O2 x (4 mols CO2/7 mols O2) = 8 L CO2 possibly formed.
Therefore, 8 L CO2 will be formed, all of the O2 will be used and some of the ethane will remain. How much will remain? That's 14 L O2 x (2 mols C2H6/7 mols O2) = 4 mols ethane used to react with the 14 L O2 which leaves 10 L ethane remaining.
ethane, 10L
To determine which gas will be leftover and the volume of that gas remaining, we need to compare the stoichiometry of the reactants and the products in the balanced chemical equation:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
From the equation, we can see that the ratio of ethane to carbon dioxide is 2:4 or 1:2. This means that for every 2 moles of ethane reacted, we will get 4 moles of carbon dioxide.
Similarly, the ratio of oxygen to carbon dioxide is 7:4. This means that for every 7 moles of oxygen reacted, we will also get 4 moles of carbon dioxide.
Since we have equal volumes of ethane and oxygen (14 L each), we can assume they have the same number of moles. Therefore, we have 2 moles of ethane and 7 moles of oxygen.
To determine the limiting reactant, compare the ratio of moles of oxygen to carbon dioxide:
7 moles O2 : 4 moles CO2
Since the ratio is 7:4, we can calculate how many moles of carbon dioxide will be produced from the 7 moles of oxygen:
(7 moles O2) x (4 moles CO2 / 7 moles O2) = 4 moles CO2
We can conclude that all 7 moles of oxygen will be completely reacted, producing 4 moles of carbon dioxide.
Now, let's calculate the mole ratio of ethane to carbon dioxide:
2 moles C2H6 : 4 moles CO2
To determine the exact amount of carbon dioxide produced, we can use the 2:4 ratio:
(2 moles C2H6) x (4 moles CO2 / 2 moles C2H6) = 4 moles CO2
This confirms that 4 moles of carbon dioxide will be produced from the 2 moles of ethane.
Since the ratio of ethane to carbon dioxide is 2:4 or 1:2, this means that ethane is the limiting reactant. It will be completely consumed, and there will be no leftover ethane.
Now, let's calculate the volume of carbon dioxide produced. Since we know that 1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP), we can calculate the volume of carbon dioxide:
(4 moles CO2) x (22.4 L / 1 mole CO2) = 89.6 L CO2
Therefore, when 14 L of ethane and 14 L of oxygen are completely reacted, 89.6 L of carbon dioxide will be produced, and there will be no leftover ethane or carbon dioxide.
To determine which gas will be leftover after the reaction and the volume of that gas remaining, we need to examine the stoichiometry of the balanced equation.
Given:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
The balanced equation tells us that 2 moles of ethane (C2H6) will react with 7 moles of oxygen (O2) to produce 4 moles of carbon dioxide (CO2) and 6 moles of water (H2O).
From the balanced equation, we can see that the ratio of ethane to oxygen is 2:7. Therefore, for every 2 moles of ethane, we need 7 moles of oxygen.
To determine which gas will be leftover, we need to find out which reactant is present in excess. If the ratio of ethane to oxygen in the mixture is different from 2:7, we can determine which one is in excess.
Using the ideal gas law equation (PV = nRT), we can analyze the moles of each gas based on their given volumes. Assuming the gases are at the same temperature and pressure, the volume can be directly proportional to the number of moles for a given gas.
Given:
Volume of ethane (C2H6) = 14 L
Volume of oxygen (O2) = 14 L
First, calculate the number of moles for each gas:
Moles of ethane = Volume of ethane (in L) × (1 mole/22.4 L) [since 1 mole of any gas at STP occupies 22.4 L]
= 14 L × (1 mole/22.4 L)
= 0.625 moles of ethane
Moles of oxygen = Volume of oxygen (in L) × (1 mole/22.4 L)
= 14 L × (1 mole/22.4 L)
= 0.625 moles of oxygen
Now, let's examine the ratio of ethane to oxygen:
Ratio of moles of ethane:moles of oxygen = 0.625 moles:0.625 moles
= 1:1
Since the ratio of the moles of ethane to oxygen is 1:1, neither of the reactants is present in excess. This means that both reactants will react completely, without leaving any excess. Therefore, there will be no leftover gas after the reaction.
To summarize, in this reaction, since both ethane and oxygen are present in equal amounts, neither of them will be leftover. The reaction will consume all the ethane and oxygen, producing 4 moles of carbon dioxide and 6 moles of water.