Calculate the volume occupied by 6.6 kg of argon gas at STP.

Answer 1

To calculate the volume occupied by 6.6 kg of argon gas at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure = 1 atm
V = volume
n = number of moles of gas
R = ideal gas constant = 0.0821 L.atm/mol.K
T = temperature = 273.15 K

First, we need to calculate the number of moles of argon gas in 6.6 kg. The molar mass of argon is approximately 39.95 g/mol.

Number of moles = mass / molar mass
Number of moles = 6600 g / 39.95 g/mol
Number of moles = 165.29 mol

Now, we can substitute the values into the ideal gas law equation:

(1 atm) V = (165.29 mol) (0.0821 L.atm/mol.K) (273.15 K)
V = 3692.38 L

Therefore, 6.6 kg of argon gas at STP would occupy approximately 3692.38 liters of volume.