How many grams of chlorine gas are present in a 150.0 liter cylinder of chlorine held at a pressure of 1.00 atm and
You didn't finish the T but use PV = nRT.
To determine the number of grams of chlorine gas present in a 150.0 liter cylinder of chlorine held at a pressure of 1.00 atm and a temperature of 25.0°C, we can use the ideal gas law equation:
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)
First, we need to convert the given temperature from Celsius to Kelvin. The conversion formula is:
K = °C + 273.15
So, T = 25.0°C + 273.15 = 298.15 K
Next, we rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the given values:
n = (1.00 atm) * (150.0 L) / (0.0821 L·atm/mol·K * 298.15 K)
Now, we can calculate the number of moles of chlorine gas:
n = 5.85 moles
Lastly, we can convert the number of moles to grams using the molar mass of chlorine gas, which is 70.91 g/mol:
grams = moles * molar mass
grams = 5.85 moles * 70.91 g/mol
grams = 414.14 grams
Therefore, there are approximately 414.14 grams of chlorine gas present in the 150.0 liter cylinder.