Butane,C4H10 , is a component of natural gas that is used as fuel for cigarette lighters.At 1.00atm and 23 degrees Celsius , how many liters of carbon dioxide are formed by the combustion of 1.00 of butane?
2C4H10+13O2 yields 8CO2+10H2o
Any help would be greatly appreciated
You have the balanced equation. Is the butane at the same P and T as CO2? Is that 1.00 liters of butane or some other unit?
Use PV = nRT. You have pressure, you have volume of butane (must be in liters), you know R, and T is 23 degrees C but don't forget to change to Kelvin. Calculate n, the number of mols of butane you start with.
Second part. Use the coefficients in the balanced equation to change mols butane to mols CO2, then use PV = nRT to calculate volume CO2. Post your work if you get stuck.
13.26
To find the volume of carbon dioxide formed by the combustion of 1.00 mole of butane, we can use the balanced chemical equation and the ideal gas law.
Step 1: Convert the temperature to Kelvin.
23 degrees Celsius = 23 + 273.15 = 296.15 K
Step 2: Use the ideal gas law, PV = nRT, to find the number of moles of carbon dioxide, assuming the volume is unknown. Rearranging the equation, we have:
n = PV / RT
The pressure (P) is given as 1.00 atm, and the temperature (T) is 296.15 K. The value of the ideal gas constant (R) is 0.0821 L.atm/mol.K.
Step 3: Calculate the number of moles of carbon dioxide formed.
n = (1.00 atm) * (V) / (0.0821 L.atm/mol.K * 296.15 K)
Since the mole ratio of butane to carbon dioxide is 2:8, and we are given 1.00 mole of butane, we expect 4.00 moles of carbon dioxide to be formed.
Step 4: Calculate the volume of carbon dioxide formed using the volume-mole relationship:
V = n * 22.4 L/mol
V = 4.00 moles * 22.4 L/mol = 89.6 L
Therefore, the volume of carbon dioxide formed by the combustion of 1.00 mole of butane is 89.6 liters.