Octane has a density of 0.692 at 20degree C . How many grams of O2 are required to burn 17.0gal of C8H18 ?
Step-by-step explanation much appreciated with as much detail as possible so I can understand. Thanks!
0.692 what? g/mL?
2C8H18 + 25O2 ==> 16CO2 + 18H2O
Convert 17.0 gallons octane to L, then to mL.
Use density (in g/mL?) to convert mL octane to grams octane.
Convert grams octane to mols. mols = grams/molar mass
Use the coefficients in the balanced equation to convert mols octane to mols O2.
Now convert mols O2 to grams. g = mols x molar mass.
To solve this problem, we need to follow these steps:
Step 1: Convert gallons to liters.
Step 2: Convert the volume of C8H18 in liters to moles.
Step 3: Write and balance the combustion equation.
Step 4: Use the balanced equation to determine the stoichiometry between C8H18 and O2.
Step 5: Calculate the number of moles of O2 required.
Step 6: Convert moles of O2 to grams.
Let's now go through each step in detail:
Step 1: Convert gallons to liters.
1 gallon is approximately equal to 3.785 liters.
So, 17.0 gallons can be converted to liters as follows:
17.0 gallons * 3.785 liters/gallon = 64.345 liters.
Step 2: Convert the volume of C8H18 in liters to moles.
To convert from volume to moles of C8H18, we need to use the ideal gas law equation:
PV = nRT, where:
P = pressure (usually at STP, which is 1 atm)
V = volume of the gas in liters
n = number of moles
R = ideal gas constant (0.0821 L * atm / (mol * K))
T = temperature in Kelvin
We are given the temperature in degrees Celsius, so we need to convert it to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 20 + 273.15 = 293.15 K
Now, we can use the ideal gas law equation to calculate the moles of C8H18:
PV = nRT
1 atm * V(L) = n * 0.0821 L * atm / (mol * K) * T(K)
Simplifying the equation:
V(L) = (n * 0.0821 * T(K)) / 1
Substituting the given values:
64.345 L = (n * 0.0821 * 293.15) / 1
Solving for n:
n = (64.345 L * 1) / (0.0821 * 293.15)
n ≈ 2.269 mol
So, we have approximately 2.269 moles of C8H18.
Step 3: Write and balance the combustion equation.
The balanced equation for the combustion of C8H18 is:
C8H18 + 12.5O2 → 8CO2 + 9H2O
Step 4: Use the balanced equation to determine the stoichiometry between C8H18 and O2.
From the balanced equation, we can see that the stoichiometric ratio between C8H18 and O2 is 1:12.5. This means that for every mole of C8H18, we need 12.5 moles of O2.
Step 5: Calculate the number of moles of O2 required.
Since we have approximately 2.269 moles of C8H18, we can now calculate the moles of O2 required:
Number of moles of O2 = 2.269 mol of C8H18 * 12.5 mol of O2/1 mol of C8H18
Number of moles of O2 ≈ 28.363 mol
Step 6: Convert moles of O2 to grams.
To convert the moles of O2 to grams, we need to use the molar mass of O2, which is 32 g/mol:
Mass of O2 = Number of moles of O2 * Molar mass of O2
Mass of O2 ≈ 28.363 mol * 32 g/mol
Mass of O2 ≈ 908.336 g
Therefore, approximately 908.336 grams of O2 are required to burn 17.0 gallons of C8H18.
To calculate the amount of grams of O2 required to burn 17.0 gallons of C8H18, we first need to understand the chemical equation for the combustion of octane (C8H18).
The balanced equation for the combustion of octane is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
This equation tells us that for every 2 moles of octane (C8H18) burned, we need 25 moles of O2. Now, let's calculate the number of moles of C8H18 in 17.0 gallons of octane.
Step 1: Convert gallons of C8H18 to liters.
Since the density of octane is given as 0.692 g/mL, we need to convert the volume from gallons to liters.
1 gallon = 3.78541 liters
So, 17.0 gallons = 17.0 * 3.78541 liters = 64.257 liters.
Step 2: Convert liters of C8H18 to grams.
To do this, we will use the density of octane.
Density = mass/volume
=> mass = density * volume
=> mass = 0.692 g/mL * 64.257 liters
=> mass = 44.491 g
Step 3: Convert grams of C8H18 to moles.
To convert grams of C8H18 to moles, we need to know the molar mass of octane (C8H18).
C8H18 has 8 carbon atoms (8*12.01 g/mol = 96.08 g/mol) and 18 hydrogen atoms (18*1.01 g/mol = 18.18 g/mol).
So, the molar mass of octane is 96.08 g/mol + 18.18 g/mol = 114.26 g/mol.
Now we can calculate the number of moles of C8H18:
moles = mass/molar mass = 44.491 g / 114.26 g/mol ≈ 0.389 mol
Step 4: Calculate the number of moles of O2 required.
From the balanced equation, we know that 2 moles of C8H18 react with 25 moles of O2.
So, for 0.389 moles of C8H18, the number of moles of O2 required can be calculated using the following proportion:
(0.389 mol C8H18) / (2 mol C8H18) = (x mol O2) / (25 mol O2)
Cross-multiplying, we get:
0.389 mol C8H18 * 25 mol O2 = 2 mol C8H18 * x mol O2
9.725 mol O2 = 2x mol O2
Dividing both sides by 2, we get:
x mol O2 = 9.725 mol O2 / 2
x mol O2 ≈ 4.863 mol O2
Step 5: Convert moles of O2 to grams of O2.
To convert moles of O2 to grams, we need to know the molar mass of O2.
O2 has 2 oxygen atoms, and the molar mass of oxygen is 16.00 g/mol.
So, the molar mass of O2 is 2 * 16.00 g/mol = 32.00 g/mol.
Now we can calculate the mass of O2:
mass = moles * molar mass = 4.863 mol O2 * 32.00 g/mol ≈ 155.416 g O2
Therefore, approximately 155.416 grams of O2 are required to burn 17.0 gallons of C8H18.