how many grams of butane must be burned in an excess of o2 gas in order to produce 7.64l of co2?
At what conditions? The volume of CO2 varies with both T and P. I'll assume STP.
2C4H10 + 13O2 ==> 8CO2 + 10H2O
mols CO2 = 7.64/22.4 = ?
Using the coefficients in the balanced equation, convert mols CO2 to mols C4H10.
Then convert mols C4H10 to grams. g = mols C4H10 x molar mass C4H10.
To determine the amount of butane needed to produce 7.64 L of CO2, we need to use the balanced chemical equation for the combustion of butane.
The balanced equation is:
C4H10 + 13/2 O2 -> 4CO2 + 5H2O
From the equation, we can see that 1 mole of butane (C4H10) reacts with 13/2 moles of O2 gas (oxygen) to produce 4 moles of CO2.
To find the amount of butane needed, we need to convert 7.64 L of CO2 into moles using the ideal gas law.
1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 L. Therefore, we can calculate the number of moles of CO2 using the following equation:
moles of CO2 = volume of CO2 (in liters) / 22.4 L
moles of CO2 = 7.64 L / 22.4 L = 0.34143 moles
According to the balanced equation, 1 mole of butane (C4H10) produces 4 moles of CO2. Therefore, the number of moles of butane required is:
moles of butane = moles of CO2 / 4
moles of butane = 0.34143 moles / 4 = 0.0853575 moles
To convert moles of butane to grams, we need to multiply by the molar mass of butane (C4H10), which is approximately 58.12 g/mol.
grams of butane = moles of butane * molar mass of butane
grams of butane = 0.0853575 moles * 58.12 g/mol = 4.95 grams
Therefore, approximately 4.95 grams of butane must be burned in an excess of O2 gas to produce 7.64 L of CO2.
To determine the number of grams of butane needed to produce 7.64 liters of CO2, we need to use stoichiometry, which involves calculating the amount of one substance based on the amount of another substance in a balanced chemical equation.
First, let's write the balanced chemical equation for the combustion of butane (C4H10) with oxygen (O2) gas to produce carbon dioxide (CO2) and water (H2O):
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen gas to produce 8 moles of CO2. Therefore, the mole ratio between butane and CO2 is 2:8 or 1:4.
To find the moles of CO2 produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (assume constant)
V = volume of CO2 (7.64 L in this case)
n = number of moles of CO2 (what we want to find)
R = ideal gas constant (constant)
T = temperature (assume constant)
Since the pressure, temperature, and volume are constant, we can rewrite the equation as:
n = (PV) / RT
Now, let's plug in the values:
P = constant pressure (assume 1 atm)
V = 7.64 L
n = ?
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = constant temperature (assume room temperature, let's say 298 K)
n = (1 atm * 7.64 L) / (0.0821 L·atm/(mol·K) * 298 K)
Simplifying the equation gives:
n ≈ 0.324 moles CO2
Since the mole ratio between butane and CO2 is 1:4, the number of moles of butane required will be four times that:
n(butane) = 4 * 0.324 moles
n(butane) ≈ 1.296 moles
Finally, we need to convert the moles of butane to grams by using the molar mass of butane:
Molar mass of butane (C4H10) = (4 * atomic mass of carbon) + (10 * atomic mass of hydrogen)
Using atomic masses in grams/mol:
Molar mass of butane = (4 * 12.01 g/mol) + (10 * 1.01 g/mol)
Molar mass of butane ≈ 58.12 g/mol
Now, we can calculate the mass of butane:
mass(butane) = n(butane) * molar mass of butane
mass(butane) ≈ 1.296 moles * 58.12 g/mol
The grammatical answer will be:
mass(butane) ≈ 75.24 grams
Therefore, approximately 75.24 grams of butane must be burned in an excess of O2 gas to produce 7.64 L of CO2.