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)+13O(g)-> 8CO2(g)+10H2O(l)
At 1.00 atm and 23C , what is the volume of carbon dioxide formed by the combustion of 3.40 grams of butane?
PV=nRT or n=PV/RT
R=0.08206
You made a typo on the equation. You omitted the 2 on O2.
mols butane = 3.40g/molar mass C4H10 = ?
Convert mols C4H10 to mols CO2 using the coefficients in the balanced equation.
Use PV = nRT to convert mols CO2 to volume in L.
Thanks so much!
To find the volume of carbon dioxide formed by the combustion of 3.40 grams of butane, we need to first calculate the number of moles of butane and then use the balanced equation to determine the moles of carbon dioxide produced. Finally, we can use the ideal gas law equation (PV = nRT) to find the volume.
Step 1: Calculate the number of moles of butane.
The molar mass of butane (C4H10) is:
4(12.01 g/mol) + 10(1.01 g/mol) = 58.12 g/mol
Therefore, the number of moles of butane is:
3.40 g / 58.12 g/mol = 0.0586 mol
Step 2: Use the balanced equation to determine the moles of carbon dioxide produced.
From the balanced equation, we can see that 2 moles of butane produce 8 moles of carbon dioxide.
So, the moles of carbon dioxide produced:
(0.0586 mol butane) x (8 mol CO2 / 2 mol butane) = 0.2344 mol carbon dioxide
Step 3: Use the ideal gas law equation to find the volume of carbon dioxide.
We are given the values for pressure (P = 1.00 atm), temperature (T = 23°C = 296K), and the value of R (0.08206 L·atm/mol·K).
Rearranging the ideal gas law equation to solve for volume:
V = (nRT) / P
Substituting the known values:
V = (0.2344 mol) x (0.08206 L·atm/mol·K) x (296K) / (1 atm)
Calculating the volume:
V = 5.89 L
Therefore, the volume of carbon dioxide formed by the combustion of 3.40 grams of butane is 5.89 liters.
To find the volume of carbon dioxide formed by the combustion of 3.40 grams of butane, we need to use the ideal gas law formula PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we need to find the number of moles of butane. We can use the molar mass of butane (C4H10) to convert the given mass (3.40 grams) to moles.
Molar mass of C4H10:
Carbon (C): 12.01 g/mol * 4 = 48.04 g/mol
Hydrogen (H): 1.01 g/mol * 10 = 10.10 g/mol
Total molar mass of butane (C4H10): 48.04 g/mol + 10.10 g/mol = 58.14 g/mol
Now, we can calculate the moles of butane:
n = mass / molar mass
n = 3.40 g / 58.14 g/mol = 0.0585 mol
Next, we need to apply stoichiometry to determine the number of moles of carbon dioxide produced. According to the balanced equation, every 2 moles of butane produce 8 moles of carbon dioxide.
Since 2 moles of butane produce 8 moles of carbon dioxide, we can set up a ratio:
2 moles of butane : 8 moles of carbon dioxide
0.0585 moles of butane : x moles of carbon dioxide
Using the ratio, we can calculate the moles of carbon dioxide:
x = (8 mol / 2 mol) * 0.0585 mol = 0.234 mol
Now, we have the number of moles of carbon dioxide produced. To find the volume of carbon dioxide, we can rearrange the ideal gas law formula and solve for V:
V = nRT / P
Plugging in the known values:
n = 0.234 mol
R = 0.08206 L∙atm/(mol∙K) (as given)
T = 23 °C = 273 K (converted to Kelvin)
P = 1.00 atm
V = (0.234 mol * 0.08206 L∙atm/(mol∙K) * 273 K) / 1.00 atm
V = 5.83 L (rounded to two decimal places)
Therefore, the volume of carbon dioxide formed by the combustion of 3.40 grams of butane at 1.00 atm and 23°C is approximately 5.83 liters.