If a tank contains 55.2 g of butane, what volume, in liters, of oxygen is needed to burn all the butane at 0.850 atm and 25 degrees Celsius? Need to show work too.
2C4H10 + 13O2 ==> 8CO2 + 10H2O
mols butane = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols butane to mols O2.
Use PV = nRT and solve for V O2 at the conditions listed.
To solve the problem, we need to use the Ideal Gas Law: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the given values to the proper units:
- Butane mass: 55.2 g
- Pressure: 0.850 atm
- Temperature: 25 degrees Celsius
The molar mass of butane (C4H10) can be calculated as follows:
- Carbon (C) molar mass: 12.01 g/mol
- Hydrogen (H) molar mass: 1.008 g/mol
- Butane molar mass = 4 * C + 10 * H = (4 * 12.01) + (10 * 1.008) = 58.12 g/mol
Next, we need to calculate the number of moles of butane:
- Number of moles = Mass / Molar mass = 55.2 g / 58.12 g/mol = 0.949 mol
Now, we can convert the temperature from Celsius to Kelvin:
- Temperature in Kelvin = 25 degrees Celsius + 273.15 = 298.15 K
At this point, we have the necessary information to proceed with the calculation.
Let's rearrange the Ideal Gas Law equation to solve for volume (V):
V = (nRT) / P
Substituting the known values:
V = (0.949 mol * 0.0821 L·atm/mol·K * 298.15 K) / 0.850 atm
After calculating the expression, we get:
V ≈ 25.63 L
Therefore, approximately 25.63 liters of oxygen is needed to burn all the butane.