Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)
At 1.00 atm and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 3.20g of butane?
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)
mols butane = grams/molar mass
Using the coefficients in the balanced equation, convert mols butane to mols CO2.
Now use PV = nRT and solve for V in liters. n = number of mols calculated in step 2.
To find the volume of carbon dioxide formed by the combustion of 3.20 g of butane, we need to use the balanced equation and the ideal gas law.
First, let's calculate the number of moles of butane:
Molar mass of butane (C4H10) = 4(12.01 g/mol) + 10(1.01 g/mol) = 58.12 g/mol
Number of moles of butane = mass / molar mass = 3.20 g / 58.12 g/mol ≈ 0.055 mol
From the balanced equation, we can see that the mole ratio between butane and carbon dioxide is 2:8 or 1:4.
So, the number of moles of carbon dioxide formed is 4 times the number of moles of butane:
Number of moles of CO2 = 4 * 0.055 mol = 0.22 mol
Now, we can use the ideal gas law to find the volume of carbon dioxide:
PV = nRT
P = 1.00 atm (given)
V = volume of CO2 (what we need to find)
n = 0.22 mol (from above)
R = 0.0821 L·atm/mol·K (ideal gas constant)
T = 23 °C = 23 + 273.15 K = 296.15 K
Plugging the values into the ideal gas law equation:
(1.00 atm) * V = (0.22 mol) * (0.0821 L·atm/mol·K) * (296.15 K)
Simplifying:
V = (0.22 mol) * (0.0821 L·atm/mol·K) * (296.15 K) / (1.00 atm)
V ≈ 5.68 L
Therefore, the volume of carbon dioxide formed by the combustion of 3.20 g of butane at 1.00 atm and 23 °C is approximately 5.68 L.
To find the volume of carbon dioxide formed by the combustion of butane, we can use the ideal gas law equation: PV = nRT.
First, we need to find the number of moles of butane (C4H10) that react. To do this, we can use the molar mass of butane.
1 mole of C4H10 = 58.12 g/mol (calculated from the atomic masses of carbon and hydrogen)
Moles of C4H10 = mass / molar mass
Moles of C4H10 = 3.20 g / 58.12 g/mol
Moles of C4H10 = 0.055 mol (approximately)
From the balanced equation, we can see that 2 moles of C4H10 react to produce 8 moles of CO2. Therefore, the number of moles of CO2 produced can be calculated as follows:
Moles of CO2 = (moles of C4H10) * (8 moles of CO2 / 2 moles of C4H10)
Moles of CO2 = 0.055 mol * (8 mol / 2 mol)
Moles of CO2 = 0.22 mol
Now, we can use the ideal gas law to find the volume of CO2.
PV = nRT
Assuming standard temperature and pressure (STP) conditions (0 °C or 273 K, and 1 atm), we can rearrange the equation to solve for the volume (V):
V = (nRT) / P
Plugging in the values:
V = (0.22 mol * 0.0821 L·atm/mol·K * 296 K) / 1 atm
V = 5.41 L (approximately)
Therefore, at 1.00 atm and 23 °C, the volume of carbon dioxide formed by the combustion of 3.20 g of butane is approximately 5.41 L.