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
If you will post your work I will find the error.
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
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.