A sample of natural gas is 80.0% CH4 and 20.0% C2H6 by mass. What is the heat from the combustion of 1.00 g of this mixture? Assume the products are CO2 and H2O.
CH4 = -74.87 and C2H6 = -84.68
Answered previously.
The delta Hs you have listed. Is that per mol or per gram? And the unit is J or kJ?
Write balanced equations for the combustion of each gas.
0.800 x 1.00 x delta H CH4 = ??
0.200 x 1.00 x delta H C2H6 = xx
?? + xx = total for mixture.
Note: If delta H is J or kJ/g the above is ok. If delta H is J or kJ/mol, you must first change the 1.00 g to mols for each gas. When you write the equation, adjust the delta Hs you have to account for the number of mols of CH4 or C2H6 if the corefficient isn't 1.
Is the oxygen not involved in this calculation of heat from combustion?
To calculate the heat from the combustion of the natural gas mixture, we need to calculate the moles of CH4 and C2H6 in the given sample.
Step 1: Calculate the moles of CH4:
To do this, we'll use the molar mass of CH4 (methane), which is 16.04 g/mol.
moles of CH4 = (mass of CH4 / molar mass of CH4)
moles of CH4 = (0.80 g / 16.04 g/mol)
Step 2: Calculate the moles of C2H6:
To do this, we'll use the molar mass of C2H6 (ethane), which is 30.07 g/mol.
moles of C2H6 = (mass of C2H6 / molar mass of C2H6)
moles of C2H6 = (0.20 g / 30.07 g/mol)
Step 3: Calculate the heat from the combustion:
Now that we have the moles of CH4 and C2H6, we can calculate the heat from the combustion. The balanced equation for the combustion of CH4 and C2H6 is:
CH4 + 2O2 -> CO2 + 2H2O
C2H6 + 7/2O2 -> 2CO2 + 3H2O
From the above balanced equations, it is clear that moles of CH4 and C2H6 will produce equal numbers of moles of CO2 and H2O.
Therefore, moles of CO2 = moles of CH4 and moles of H2O = 2 * moles of CH4.
heat from the combustion = (moles of CO2 * enthalpy of CO2) + (moles of H2O * enthalpy of H2O)
heat from the combustion = (moles of CH4 * enthalpy of CO2) + (2 * moles of CH4 * enthalpy of H2O)
Let's calculate the heat from the combustion:
heat from the combustion = (0.80 g / 16.04 g/mol) * (-74.87 kJ/mol) + (2 * (0.80 g / 16.04 g/mol)) * (-84.68 kJ/mol)
Now, let's go step by step to solve this equation:
Step 4: Calculate the moles of CH4:
moles of CH4 = (0.80 g / 16.04 g/mol)
moles of CH4 = 0.04988 mol (approximately 0.050 mol)
Step 5: Calculate the moles of CO2:
moles of CO2 = moles of CH4 (as mentioned earlier)
moles of CO2 = 0.050 mol
Step 6: Calculate the moles of H2O:
moles of H2O = 2 * moles of CH4 (as mentioned earlier)
moles of H2O = 2 * 0.050 mol
moles of H2O = 0.100 mol
Step 7: Calculate the heat from the combustion:
heat from the combustion = (0.050 mol * -74.87 kJ/mol) + (2 * 0.050 mol * -84.68 kJ/mol)
heat from the combustion = -3.7435 kJ + (-8.468 kJ)
heat from the combustion = -12.2115 kJ
Therefore, the heat from the combustion of 1.00 g of this mixture is approximately -12.21 kJ.
To calculate the heat from the combustion of the given mixture, we need to determine the amount of each component that will react and then calculate the heat released from the combustion of each component separately. Finally, we will sum up the individual heats to obtain the total heat released.
Step 1: Calculate the mass of CH4 and C2H6 in the given 1.00 g mixture:
Mass of CH4 = 80.0% of 1.00 g = 0.80 g
Mass of C2H6 = 20.0% of 1.00 g = 0.20 g
Step 2: Calculate the moles of CH4 and C2H6:
Molar mass of CH4 = 12.01 g/mol (C) + 4(1.01 g/mol) (H) = 16.04 g/mol
Moles of CH4 = Mass of CH4 / Molar mass of CH4 = 0.80 g / 16.04 g/mol ≈ 0.050 mol
Molar mass of C2H6 = 2(12.01 g/mol) (C) + 6(1.01 g/mol) (H) = 30.07 g/mol
Moles of C2H6 = Mass of C2H6 / Molar mass of C2H6 = 0.20 g / 30.07 g/mol ≈ 0.007 mol
Step 3: Calculate the heat released from the combustion of each component using the given enthalpy values (-74.87 kJ/mol for CH4 and -84.68 kJ/mol for C2H6):
Heat released from CH4 combustion = Moles of CH4 × Enthalpy of combustion of CH4
= 0.050 mol × (-74.87 kJ/mol)
= -3.74 kJ
Heat released from C2H6 combustion = Moles of C2H6 × Enthalpy of combustion of C2H6
= 0.007 mol × (-84.68 kJ/mol)
= -0.59 kJ
Step 4: Sum up the individual heats to obtain the total heat released:
Total heat released = Heat released from CH4 combustion + Heat released from C2H6 combustion
= -3.74 kJ + (-0.59 kJ)
= -4.33 kJ
Therefore, the heat released from the combustion of 1.00 g of the given natural gas mixture is approximately -4.33 kJ.