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 1.40 g of butane?
Express your answer with the appropriate units.
NOTE: I got 1.676L, but it says that it is a wrong answer :(
Why didn't you show how you obtained that 1.676 L. I could have taken one look at it and told you where you went wrong.
How many mols is 1.4 grams butane. That's mols = g/molar mass = 1.40/58 = approx 0.024.
Convert mols C4H10 to mols CO2 produced. That's
0.024 mols C4H10 x (8 mols CO2/2 mol C4H10) = 0.024 x 4 = about 0.096 mols CO2. Plug that into
PV = nRT
1*V = 0.096 x 0.08206*(23+273)
Solve for V in liters.
Post your work if you get stuck.
Well, it seems like your answer went up in smoke! Let's take a closer look at the problem.
First, we need to find the moles of butane we have. The molar mass of butane (C4H10) is 58.12 g/mol. So, 1.40 g of butane is equal to:
1.40 g / 58.12 g/mol = 0.0241 mol
According to the balanced equation, 2 moles of butane produce 8 moles of carbon dioxide. So, we can set up a ratio:
0.0241 mol C4H10 / 2 mol C4H10 = x mol CO2 / 8 mol CO2
Cross-multiplying and solving for x, we find:
x = (0.0241 mol C4H10 * 8 mol CO2) / 2 mol C4H10 = 0.0964 mol CO2
Next, we need to convert moles of CO2 to volume using the ideal gas law. At 1.00 atm and 23 °C, the volume of a gas can be determined using the formula:
V = (nRT) / P
Where:
V = volume in liters
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
P = pressure in atmospheres
The value of R is 0.0821 L·atm/(mol·K).
Converting 23 °C to Kelvin, we get:
T = 23 °C + 273.15 = 296.15 K
Plugging in the values, we have:
V = (0.0964 mol CO2 * 0.0821 L·atm/(mol·K) * 296.15 K) / 1.00 atm = 2.19 L
So, the correct answer is 2.19 liters of carbon dioxide.
Keep in mind that significant figures might have affected your answer, so make sure to round according to the given significant figures in the problem.
To find the volume of carbon dioxide formed by the combustion of 1.40 g of butane, we can use the balanced equation to determine the molar ratio between butane and carbon dioxide. Then, we can use the ideal gas law to calculate the volume.
1. Calculate the number of moles of butane using its molar mass:
molar mass of butane (C4H10) = 4 * atomic mass of carbon + 10 * atomic mass of hydrogen
= 4 * 12.01 g/mol + 10 * 1.01 g/mol
= 58.12 g/mol
moles of butane = mass of butane / molar mass of butane
= 1.40 g / 58.12 g/mol
= 0.0241 mol
2. Use the balanced equation to determine the mole ratio between butane and carbon dioxide:
According to the balanced equation, 2 moles of butane produce 8 moles of carbon dioxide.
moles of carbon dioxide = moles of butane * (8 moles CO2 / 2 moles butane)
= 0.0241 mol * (8 mol CO2 / 2 mol butane)
= 0.0964 mol CO2
3. Apply the ideal gas law to calculate the volume of carbon dioxide:
The ideal gas law is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.
Rearranging the equation, V = (nRT) / P
Assuming constant pressure (1.00 atm) and temperature (23 °C = 23 + 273.15 = 296.15 K), and using the ideal gas constant (R = 0.0821 L·atm/(mol·K)), we can calculate the volume.
V = (0.0964 mol CO2 * 0.0821 L·atm/(mol·K) * 296.15 K) / 1.00 atm
= 2.263 L
Therefore, the correct volume of carbon dioxide formed by the combustion of 1.40 g of butane is 2.263 L.
To find the volume of carbon dioxide formed by the combustion of 1.40 g of butane, we can use the balanced equation and the ideal gas law.
First, we need to calculate the number of moles of butane. We can do this by dividing the given mass by the molar mass of butane.
The molar mass of butane (C4H10) is calculated by adding up the molar masses of carbon (C) and hydrogen (H).
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of butane (C4H10) = (4 * Molar mass of C) + (10 * Molar mass of H)
= (4 * 12.01 g/mol) + (10 * 1.008 g/mol)
= 58.12 g/mol
Next, we can calculate the number of moles of butane:
Number of moles of butane = Mass of butane / Molar mass of butane
= 1.40 g / 58.12 g/mol
≈ 0.02409 mol (rounded to 5 decimal places)
From the balanced equation, we can see that 2 moles of butane produce 8 moles of carbon dioxide (CO2).
Using the stoichiometry, we can determine the moles of CO2 produced:
Number of moles of CO2 = (Number of moles of butane) * (8 moles of CO2 / 2 moles of butane)
= (0.02409 mol) * (8 mol CO2 / 2 mol butane)
= 0.09636 mol
Now, we can use the ideal gas law to find the volume of carbon dioxide at 1.00 atm and 23 °C.
Ideal Gas Law: PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 23 °C + 273.15
= 296.15 K
Now, we can substitute the values into the ideal gas law:
(1.00 atm) * (V) = (0.09636 mol) * (0.0821 L·atm/(mol·K)) * (296.15 K)
Simplifying the equation:
V = (0.09636 mol) * (0.0821 L·atm/(mol·K)) * (296.15 K) / (1.00 atm)
V ≈ 2.376 L
Therefore, the volume of carbon dioxide formed by the combustion of 1.40 g of butane at 1.00 atm and 23 °C is approximately 2.376 L.