Final answer:
13.0 liters of Chlorine gas (Cl₂) at STP contains 0.58 moles of Cl₂. The calculation uses Avogadro's law and the concept of molar volume, where 1 mole of gas occupies 22.4 liters at STP.
Step-by-step explanation:
To determine the number of moles of Chlorine gas (Cl₂) in 13.0 liters at Standard Temperature and Pressure (STP), we can use Avogadro's law and the concept of molar volume. Avogadro's law states that one mole of any gas at STP occupies a volume of 22.4 liters. Therefore, we can set up a simple proportion, since the molar volume at STP is a constant:
- 1 mole Cl₂ / 22.4 L = x moles Cl₂ / 13.0 L
Solving for x gives us x = 13.0 L * (1 mole Cl₂/22.4 L) = 0.58 moles Cl₂. This calculation does not require us to use the molar mass of Cl₂ or convert to the number of molecules, thus making it straightforward and efficient.
Remember that when you are presented with a volume of gas at STP and you're asked to find the amount in moles, the molar volume of 22.4 L/mol is essential for your calculations.