What is the volume of 26.311g of butane gas (C4H10), at 22.7°C and 125.1 kPa? Round your answer to the nearest thousandth.
I had gotten 0.009
I get 8.9 L
not sure what your units are. What did you do?
0.009 what? liters? gallons? I think you used R in L*atm/mol*K but used P in kPa. And that's not right. PV = nRT.
125.1 x V = (26.31g/58)*8.314*295.85 = ? or
(125.1/101.325) x V = (26.31/58)*0.08205*295.85 = ?
Should be the same number for each.
To calculate the volume of butane gas, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kilopascals, kPa)
V = volume (in liters, L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K or 8.31 J/mol·K)
T = temperature (in Kelvin, K)
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 22.7°C + 273.15 = 295.85 K
Next, we need to calculate the number of moles of butane gas using the given mass and the molar mass of butane.
The molar mass of butane (C4H10) can be calculated by adding up the atomic masses of carbon and hydrogen:
(4 × atomic mass of carbon) + (10 × atomic mass of hydrogen)
(4 × 12.01 g/mol) + (10 × 1.008 g/mol) = 58.12 g/mol
Now, calculate the number of moles using the mass of butane:
n = mass / molar mass
n = 26.311 g / 58.12 g/mol ≈ 0.453 mol
Now, substitute the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (0.453 mol × 0.0821 L·atm/mol·K × 295.85 K) / 125.1 kPa
V ≈ 0.008828 L
Rounding to the nearest thousandth, the volume of 26.311g of butane gas at 22.7°C and 125.1 kPa is approximately 0.009 L.