How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0.0 °C?
its 474g...your answer is so very wrong!!
Use PV = nRT, solve for n = moles and convert moles to grams.
70x^2 ClNa2
To find the number of grams of chlorine gas present in the cylinder, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given temperature from Celsius to Kelvin:
T (in Kelvin) = T (in °C) + 273.15
T (in Kelvin) = 0.0 °C + 273.15
T (in Kelvin) = 273.15 K
Next, substitute the given values into the equation:
P = 1.00 atm
V = 150.0 liters
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
Now we can rearrange the equation to solve for the number of moles (n):
n = PV / RT
n = (1.00 atm) * (150.0 liters) / (0.0821 L·atm/(mol·K)) * (273.15 K)
n ≈ 6.44 moles
Finally, we need to convert moles to grams using the molar mass of chlorine gas, which is approximately 70.90 g/mol. Multiply the number of moles by the molar mass:
Number of grams = number of moles * molar mass
Number of grams = 6.44 moles * 70.90 g/mol
Number of grams ≈ 456 grams
Therefore, there are approximately 456 grams of chlorine gas present in the 150.0 liter cylinder at a pressure of 1.00 atm and 0.0 °C.
using PV=nRT...
n=PV/RT
= 1 atm*150L/0.08206 (atm*L/mol*K)
= 150/22.40238 mol
= 6.71 mol Cl
THEN...
6.71 mol Cl x 35.45gCl/1mol Cl = 237 g Cl