The complete combustion of octane, C8H18, the main component of gasoline, proceeds as follows.
2 C8H18(l) + 25 O2(g) �¨ 16 CO2(g) + 18 H2O(g)
Octane has a density of 0.692 g/mL at 20�‹C. How many grams of O2 are required to burn 26.5 gal of C8H18?
i got 1.22e5 but it is wrong
12 years ago
12 years ago
i cannot find my exact work but i converted 26.5 gal C8H18 to ml then used the density to get grams then converted grams to moles of C8H18 then to moles of )2 then grams of O2
1 year ago
To find the number of grams of O2 required to burn 26.5 gallons of C8H18, we can follow these steps:
Step 1: Convert the volume of C8H18 from gallons to milliliters.
Since 1 gallon is equal to approximately 3785.41 mL, we can calculate the volume of C8H18 in milliliters:
26.5 gal * 3785.41 mL/gal = 99949.365 mL
Step 2: Convert the volume of C8H18 from milliliters to grams.
To do this, we need to use the given density of octane: 0.692 g/mL.
The mass of C8H18 can be calculated as follows:
99949.365 mL * 0.692 g/mL = 69184.475 g
Step 3: Calculate the molar mass of C8H18.
The molar mass of octane (C8H18) can be calculated by summing the atomic masses:
(8 * atomic mass of C) + (18 * atomic mass of H)
Using the atomic masses from the periodic table, we find:
(8 * 12.01 g/mol) + (18 * 1.01 g/mol) = 114.23 g/mol
Step 4: Calculate the number of moles of C8H18.
The number of moles can be calculated using the formula:
moles = mass / molar mass
In this case, the number of moles is:
69184.475 g / 114.23 g/mol = 605.8253 mol
Step 5: Determine the stoichiometry ratio of O2 to C8H18.
From the balanced chemical equation, we know that 2 moles of C8H18 react with 25 moles of O2.
Thus, the stoichiometry ratio of O2 to C8H18 is 25/2.
Step 6: Calculate the number of moles of O2 required.
To find the number of moles of O2 required, we multiply the number of moles of C8H18 by the stoichiometry ratio:
605.8253 mol * (25/2) = 7585.31625 mol
Step 7: Convert the number of moles of O2 to grams.
To convert moles to grams, we need to use the molar mass of O2, which is approximately 32 g/mol.
The mass of O2 can be calculated as follows:
7585.31625 mol * 32 g/mol = 242097.000 g
Therefore, the correct answer is 242097 grams of O2 are required to burn 26.5 gallons of C8H18.