A liquid fuel mixture contains 31.45 % hexane (C6H14), 16.15 % heptane (C7H16), and the rest octane (C8H18).
What maximum mass of carbon dioxide is produced by the complete combustion of 13.0 kg of this fuel mixture?
I think the easiest way to solve this is to calculate the amount of CO2 from EACH hydrocarbon (as if the hydrocarbon was pure), then see what contribution came from that particular hydrocarbon, as follows:
hexane is C6H14 = 86 and 31.45%
heptane is C7H16 = 100 and 16.15%
octane is C8H18 = 114 and 100-31.45-16.15 = 52.40
2C6H14 + 19O2 ==> 12CO2 + 14H2O
mols C6H14 = g/molar mass = 13,000/86 = approximately 151 (this is a close estimate. You should recalculate ALl on this post to get a better number) since all of the calculations that follow are close estimates.
mols CO2 formed = 151 mols C6H14 x (12 mols CO2/2 mols(C6H14) = about 907 and 907 x 31.45% = about 28,800 mols and that x 44 g/mol = about 12,500 g CO2 from the hexane.
Now you do the same thing for heptane and octane but the combustion equation for heptane and octane will be different. Then you add the grams from each to find the total grams CO2 produced. Post your work if you get stuck.
Please how did u get the 86?
Ah, fuel chemistry, a combustibly delightful topic! To determine the maximum mass of carbon dioxide produced, we'll need to find the amount of carbon in the fuel mixture and then calculate the resulting CO2.
Let's start by determining the masses of each component in the fuel mixture:
- Mass of hexane (C6H14): 31.45% of 13.0 kg = 4.0885 kg
- Mass of heptane (C7H16): 16.15% of 13.0 kg = 2.0995 kg
- Mass of octane (C8H18): The rest! 13.0 kg - (4.0885 kg + 2.0995 kg) = 6.8120 kg
Now, let's find the moles of each component using their respective molar masses:
- Moles of hexane: 4.0885 kg / 86.18 g/mol = 47.4 mol
- Moles of heptane: 2.0995 kg / 100.21 g/mol = 20.9 mol
- Moles of octane: 6.8120 kg / 114.23 g/mol = 59.7 mol
Next, let's determine the moles of carbon in each component:
- Moles of carbon in hexane: 47.4 mol × 6 mol carbon/mol hexane = 284.4 mol carbon
- Moles of carbon in heptane: 20.9 mol × 7 mol carbon/mol heptane = 146.3 mol carbon
- Moles of carbon in octane: 59.7 mol × 8 mol carbon/mol octane = 477.6 mol carbon
Finally, let's calculate the mass of carbon dioxide produced from the combustion of each component:
- Mass of carbon dioxide from hexane: 284.4 mol carbon × (44.01 g/mol CO2 / 12.01 g/mol carbon) = 1050.6 g CO2
- Mass of carbon dioxide from heptane: 146.3 mol carbon × (44.01 g/mol CO2 / 12.01 g/mol carbon) = 537.5 g CO2
- Mass of carbon dioxide from octane: 477.6 mol carbon × (44.01 g/mol CO2 / 12.01 g/mol carbon) = 1761.4 g CO2
Now, let's sum up the masses of carbon dioxide produced from each component:
- Mass of carbon dioxide produced: 1050.6 g + 537.5 g + 1761.4 g = 3349.5 g
And finally, converting grams to kilograms:
- Maximum mass of carbon dioxide produced: 3349.5 g / 1000 = 3.35 kg
So, the maximum mass of carbon dioxide produced by the complete combustion of 13.0 kg of this fuel mixture is approximately 3.35 kg.
To find the maximum mass of carbon dioxide produced by the complete combustion of the fuel mixture, we need to calculate the moles of each component and then use the coefficients in the balanced equation to determine the moles of carbon dioxide produced.
Step 1: Calculate the moles of each component.
First, we need to calculate the moles of hexane (C6H14), heptane (C7H16), and octane (C8H18) in the fuel mixture.
Molar mass of hexane (C6H14):
Carbon: 6 * 12.01 g/mol = 72.06 g/mol
Hydrogen: 14 * 1.01 g/mol = 14.14 g/mol
Total molar mass of hexane = 72.06 + 14.14 = 86.20 g/mol
Moles of hexane = (31.45/100) * (13.0 kg) / (86.20 g/mol) = 5.39 mol
Similarly,
Moles of heptane = (16.15/100) * (13.0 kg) / (100.21 g/mol) = 1.32 mol
Moles of octane = (100 - 31.45 - 16.15)/100 * (13.0 kg) / (114.23 g/mol) = 2.60 mol
Step 2: Use the balanced equation to determine the moles of carbon dioxide produced.
The balanced equation for the complete combustion of a hydrocarbon fuel can be represented as follows:
CnHm + (n + m/4) O2 -> n CO2 + (m/2) H2O
From the balanced equation, we can see that 1 mole of hydrocarbon produces n moles of carbon dioxide.
For hexane (C6H14), n = 6.
Moles of carbon dioxide produced from hexane = 5.39 mol * 6 = 32.34 mol
For heptane (C7H16), n = 7.
Moles of carbon dioxide produced from heptane = 1.32 mol * 7 = 9.24 mol
For octane (C8H18), n = 8.
Moles of carbon dioxide produced from octane = 2.60 mol * 8 = 20.80 mol
Step 3: Calculate the mass of carbon dioxide produced.
To calculate the mass of carbon dioxide produced, we need to multiply the moles of carbon dioxide produced by its molar mass.
Molar mass of carbon dioxide (CO2): 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygen) = 44.01 g/mol
Mass of carbon dioxide produced = (32.34 + 9.24 + 20.80) mol * 44.01 g/mol = 4009 g = 4.01 kg
Therefore, the maximum mass of carbon dioxide produced by the complete combustion of 13.0 kg of this fuel mixture is 4.01 kg.
To calculate the maximum mass of carbon dioxide produced by the complete combustion of the fuel mixture, we need to determine the number of moles of each component and then use stoichiometry to find the amount of carbon dioxide produced.
First, let's calculate the mass of each component in the fuel mixture:
Mass of hexane (C6H14) = 31.45% of 13.0 kg
= 0.3145 * 13.0 kg
= 4.0835 kg
Mass of heptane (C7H16) = 16.15% of 13.0 kg
= 0.1615 * 13.0 kg
= 2.1005 kg
Mass of octane (C8H18) = (100% - 31.45% - 16.15%) of 13.0 kg
= (100% - 47.6%) of 13.0 kg
= 52.4% of 13.0 kg
= 0.524 * 13.0 kg
= 6.812 kg
Next, we need to convert the masses of each component to moles:
Molar mass of hexane (C6H14) = (6 * atomic mass of carbon) + (14 * atomic mass of hydrogen)
= (6 * 12.01 g/mol) + (14 * 1.01 g/mol)
= 86.18 g/mol
Number of moles of hexane = Mass of hexane / Molar mass of hexane
= 4.0835 kg / (86.18 g/mol / 1000 g/kg)
= 47.437 mol
Molar mass of heptane (C7H16) = (7 * atomic mass of carbon) + (16 * atomic mass of hydrogen)
= (7 * 12.01 g/mol) + (16 * 1.01 g/mol)
= 100.2 g/mol
Number of moles of heptane = Mass of heptane / Molar mass of heptane
= 2.1005 kg / (100.2 g/mol / 1000 g/kg)
= 20.99 mol
Molar mass of octane (C8H18) = (8 * atomic mass of carbon) + (18 * atomic mass of hydrogen)
= (8 * 12.01 g/mol) + (18 * 1.01 g/mol)
= 114.23 g/mol
Number of moles of octane = Mass of octane / Molar mass of octane
= 6.812 kg / (114.23 g/mol / 1000 g/kg)
= 59.64 mol
Now, we can determine the balanced equation for the combustion of each component:
C6H14 + 19/2 O2 -> 6 CO2 + 7 H2O
C7H16 + 23/2 O2 -> 7 CO2 + 8 H2O
C8H18 + 25/2 O2 -> 8 CO2 + 9 H2O
Using the coefficients in the balanced equations, we can calculate the number of moles of CO2 produced from each component:
Number of moles of CO2 from hexane = 6/1 * Number of moles of hexane
= 6/1 * 47.437 mol
= 284.62 mol
Number of moles of CO2 from heptane = 7/1 * Number of moles of heptane
= 7/1 * 20.99 mol
= 146.93 mol
Number of moles of CO2 from octane = 8/1 * Number of moles of octane
= 8/1 * 59.64 mol
= 477.12 mol
Finally, we can calculate the total number of moles of CO2 produced:
Total number of moles of CO2 = Number of moles of CO2 from hexane + Number of moles of CO2 from heptane + Number of moles of CO2 from octane
= 284.62 mol + 146.93 mol + 477.12 mol
= 908.67 mol
To calculate the mass of carbon dioxide produced, we need to multiply the total number of moles of CO2 by the molar mass of CO2:
Molar mass of CO2 = (1 * atomic mass of carbon) + (2 * atomic mass of oxygen)
= (1 * 12.01 g/mol) + (2 * 16.00 g/mol)
= 44.01 g/mol
Mass of carbon dioxide produced = Total number of moles of CO2 * Molar mass of CO2
= 908.67 mol * 44.01 g/mol
= 39,989.35 g
= 39.99 kg
Therefore, the maximum mass of carbon dioxide produced by the complete combustion of 13.0 kg of this fuel mixture is 39.99 kg.