Convert from 5.6 moles of Cl2 (g) to liters at STP

thank you

To convert moles of a gas to liters at Standard Temperature and Pressure (STP), you can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)

At STP, the pressure is 1 atmosphere (atm) and the temperature is 273.15 Kelvin (K).

First, we need to find the volume (V). Rearranging the ideal gas law equation, we have:

V = (nRT) / P

Now, let's plug in the values:

n = 5.6 moles
R = 0.0821 L·atm/mol·K
T = 273.15 K
P = 1 atm

V = (5.6 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
V = 123.92 liters

Therefore, 5.6 moles of Cl2 (g) is equivalent to 123.92 liters at STP.

L x (1 mol/22.4L) = ?