How many grams of methane gas were used to fill a baloon to a volume of 3.0 L at STP?
Use PV = nRT and solve for n, then m = grams/molar mass and solve for grams.
To calculate the number of grams of methane gas used to fill a balloon to a volume of 3.0 L at STP, we need to use the ideal gas law equation, which is given as:
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)
At STP (Standard Temperature and Pressure), the pressure is 1 atm, and the temperature is 273 Kelvin.
Now, we can rearrange the equation to solve for n:
n = PV / RT
Plugging in the values:
P = 1 atm
V = 3.0 L
R = 0.0821 L.atm/mol.K
T = 273 K
n = (1 atm) * (3.0 L) / (0.0821 L.atm/mol.K * 273 K)
n = 0.1321 mol
Since the molar mass of methane is approximately 16.04 g/mol, we can calculate the mass using:
Mass = number of moles * molar mass
Mass = 0.1321 mol * 16.04 g/mol
Mass ≈ 2.12 grams
Therefore, approximately 2.12 grams of methane gas were used to fill the balloon to a volume of 3.0 L at STP.
To find the number of grams of methane gas used to fill a balloon to a volume of 3.0 L at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
At STP, the pressure is 1 atm and the temperature is 273 K. Since we are given the volume (V = 3.0 L) and want to find the number of grams (n), we need to rearrange the equation:
n = PV / RT
Before substituting the values, we need to convert the volume from liters to moles. For this, we'll use the ideal gas law:
PV = nRT
Substituting the known values:
(1 atm) (3.0 L) = n (0.0821 L·atm/(mol·K)) (273 K)
Simplifying:
3.0 atm·L = n (22.41 L·atm/(mol·K))
Now, if we divide both sides of the equation by (22.41 L·atm/(mol·K)), we can find the number of moles (n):
n = (3.0 atm·L) / (22.41 L·atm/(mol·K))
n ≈ 0.134 mol
Now that we know the number of moles, we can convert it to grams using the molar mass of methane (CH4): 16.04 g/mol.
m = n × M
Substituting the values:
m = (0.134 mol) × (16.04 g/mol)
m ≈ 2.15 g
Therefore, approximately 2.15 grams of methane gas were used to fill the balloon to a volume of 3.0 L at STP.