Given the reaction 2C2H6 + 7O2 → 4CO2 + 6H2O ∆H = -1416 kJ/mol C2H6
a. How many liters of Carbon Dioxide would be produced if 16.00 L of ethane, (C2H6), were burnt (all at STP)?
b. How many liters of water vapor would also be produced?
c. How many liters of Oxygen would have been used?
d. How many kJ of heat would have been released?
a.
16L C2H6 x (4 mols CO2/2 mols C2H6) = ? mols CO2.
b and c are done the same way.
d.
1416 kJ/mol x [(16 L) x (1 mol/22.4L)]
To answer these questions, we need to use the balanced chemical equation and the given information.
a. To determine the number of liters of carbon dioxide produced, we need to use the stoichiometry of the balanced equation. Let's first find the moles of C2H6 in 16.00 L at STP (Standard Temperature and Pressure).
At STP, 1 mole of gas occupies 22.4 L.
Therefore, the number of moles of C2H6 in 16.00 L is:
moles of C2H6 = (16.00 L) / (22.4 L/mol) = 0.71 mol
According to the stoichiometry of the balanced equation, 2 moles of C2H6 produce 4 moles of CO2.
So, the number of moles of CO2 produced is:
moles of CO2 = 4 * 0.71 mol = 2.84 mol
Now, we can convert moles of CO2 to liters at STP:
liters of CO2 = (moles of CO2) * (22.4 L/mol) = 2.84 mol * 22.4 L/mol = 63.6 L
Therefore, 16.00 L of C2H6 would produce 63.6 L of CO2.
b. Using the same approach as in part a, we can find the number of liters of water vapor produced.
According to the stoichiometry of the balanced equation, 2 moles of C2H6 produce 6 moles of H2O.
So, the number of moles of H2O produced is:
moles of H2O = 6 * 0.71 mol = 4.26 mol
Now, we can convert moles of H2O to liters at STP:
liters of H2O = (moles of H2O) * (22.4 L/mol) = 4.26 mol * 22.4 L/mol = 95.5 L
Therefore, 16.00 L of C2H6 would produce 95.5 L of water vapor.
c. To find the number of liters of oxygen used, we need to consider the stoichiometry of the balanced equation.
According to the stoichiometry, 2 moles of C2H6 require 7 moles of O2.
So, the number of moles of O2 needed is:
moles of O2 = (7/2) * 0.71 mol = 2.47 mol
Now we can convert moles of O2 to liters at STP:
liters of O2 = (moles of O2) * (22.4 L/mol) = 2.47 mol * 22.4 L/mol = 55.33 L
Therefore, 16.00 L of C2H6 would require 55.33 L of O2.
d. The enthalpy change (∆H) for the reaction is given as -1416 kJ/mol C2H6.
To find the amount of heat released, we need to calculate the moles of C2H6 burnt.
Using the same approach as in part a, we found that 16.00 L of C2H6 corresponds to 0.71 mol.
So, the heat released for 0.71 mol of C2H6 burned is:
Heat released = ∆H * moles of C2H6 = -1416 kJ/mol * 0.71 mol = -1006.36 kJ
Therefore, 16.00 L of C2H6 would release approximately 1006.36 kJ of heat.
To answer these questions, we need to use the stoichiometry of the reaction and determine the appropriate conversion factors. Here's how we can calculate each part:
a. To calculate the volume of carbon dioxide produced, we first need to know the mole ratio between ethane (C2H6) and carbon dioxide (CO2) from the balanced equation. According to the reaction equation, we have 2 moles of C2H6 producing 4 moles of CO2.
Given that 2 moles of C2H6 yield 4 moles of CO2, we can set up the following conversion factors:
1 mol C2H6 = 4 mol CO2
16.00 L C2H6 * (1 mol C2H6 / 22.4 L C2H6) * (4 mol CO2 / 2 mol C2H6) = X L CO2
Simplifying, we find:
X = (16.00 * 4) / 2 = 32.00 L CO2
Therefore, 32.00 liters of carbon dioxide would be produced when burning 16.00 liters of ethane at STP.
b. Similar to part (a), we need to determine the mole ratio between ethane (C2H6) and water (H2O) from the balanced equation. According to the reaction equation, we have 2 moles of C2H6 producing 6 moles of H2O.
Using the same steps as before, we can calculate the volume of water vapor produced:
16.00 L C2H6 * (1 mol C2H6 / 22.4 L C2H6) * (6 mol H2O / 2 mol C2H6) = Y L H2O
Simplifying, we find:
Y = (16.00 * 6) / 2 = 48.00 L H2O
Therefore, 48.00 liters of water vapor would be produced when burning 16.00 liters of ethane at STP.
c. To determine the volume of oxygen (O2) used in the reaction, we need to use the stoichiometry again. According to the reaction equation, 2 moles of C2H6 react with 7 moles of O2.
Setting up the conversion factors:
16.00 L C2H6 * (1 mol C2H6 / 22.4 L C2H6) * (7 mol O2 / 2 mol C2H6) = Z L O2
Simplifying, we find:
Z = (16.00 * 7) / 2 = 56.00 L O2
Therefore, 56.00 liters of oxygen would be used when burning 16.00 liters of ethane at STP.
d. To calculate the heat released (in kJ), we use the given enthalpy change (∆H) from the reaction. The enthalpy change for the combustion of 1 mole of ethane is -1416 kJ/mol C2H6.
Since we have burned 16.00 liters of ethane, which can be converted to moles using the molar volume at STP, we can calculate the heat released:
16.00 L C2H6 * (1 mol C2H6 / 22.4 L C2H6) * (-1416 kJ / 1 mol C2H6) = W kJ
Simplifying, we find:
W = (16.00 * -1416) / 22.4 = -1011.43 kJ
Therefore, approximately -1011.43 kJ of heat would have been released when burning 16.00 liters of ethane. The negative sign indicates that the reaction is exothermic, meaning it releases heat.