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# what is the volume of 79.3 g of Cl2 at STP?

### 4 Answers

- Lexi RLv 71 decade agoFavorite Answer
At STP 1 mole of any gas occupies a volume of 22.4 L

moles Cl2 = mass / molar mass

= 79.3 g / 70.9 g/mol

= 1.12 mol

Volume = 1.12 mol x 22.4 L/mol

= 25.1 L (3 sig fig)

PV = nRT should be used for any conditions other then STP (which is 1 atm and 0 deg C (273.15 K) It can be used for STP as well, but since we know volume of 1 mole ideal gas = 22.4 L at STP it is not essential.

- VixLv 51 decade ago
1 mol of Cl2 = 70.9 g

1 mol of Cl2 = 22.4 mol

73.9 g of Cl2 * 1 mol of Cl2 / 70.9 g of Cl2 * 22.4 L of Cl2/1 mol of Cl2=

23.3 L of Cl2

- MacDLv 41 decade ago
use the ideal gas equation.

PV = nRT

V = (nRT)/p

STP= T = 273.15 K ; P = 1 atm.

n= mass/ molar mass

= 79.3/35.45= 2.237moles

V = [(2.237moles)(0.082057)(273.15 K)] / 1 atm

=50.14 L(liters)

i don't know where you had that formula vix but it is wrong.

Source(s): chemistry undergrad