If there are 1.4 x 103 mg of chlorine gas at STP, how many liters does this occupy?

Question

Struggling with this question

Answer 1

When a gas is at STP (standard temperature and pressure) there is no need to use the ideal gas law equation (although you may). All you need to know is that at STP, 1 mole of an ideal gas = 22.4 liters.

molar mass Cl2 = 70.9 g/mol

1.4x10³ mg x 1 g / 1000 mg = 1.4 g

moles Cl2 = 1.4 g x 1 mol / 70.9 g = 0.0197 moles Cl2

literes = 0.0197 moles x 22.4 liters/mol = 0.442 liters

Answer 2

The ideal gas law states PV=nRT

P-Pressure (atm)

V-Volume (L)

n-Moles (mol)

R-Gas Constant (L*atm/mol*K)

T-Temperature(K)

STP stands for Standard Temperature and Pressure which 273.15K and 1atm, respectively.

Chlorine is a diatomic molecule, therefore Chlorine gas exists as Cl2 (70.91 g/mol)

PV = nRT

Solve for V: V=nRT / P

n= 1.4e3 mg Cl2 * (1g / 1000mg) * (1 mol Cl2 / 70.91 g Cl2) = 0.01974 mol Cl2

V = [(0.01974 mol Cl2) * (0.0821 L*atm / mol*K) * (273.15 K)] / (1 atm)

V = 0.443 L