Final answer:
To find the volume of chlorine gas from the decomposition of 85.10 g PCl₅, calculate the moles of PCl₅, use stoichiometry to get moles of Cl₂, and apply the ideal gas law to get 9.164 liters of Cl₂ at STP.
Step-by-step explanation:
To determine the volume of chlorine gas (Cl₂) that forms from the decomposition of 85.10 g of PCl₅, we first need to use the balanced chemical equation for the decomposition of PCl₅:
PCl₅(g) → PCl₃(g) + Cl₂(g)
Then we calculate the moles of PCl₅ using its molar mass and use stoichiometry to find the moles of Cl₂ produced. The molar mass of PCl₅ is approximately 208.24 g/mol. The calculation for moles of PCl₅ is:
Moles of PCl₅ = 85.10 g / 208.24 g/mol = 0.4087 mol
According to the stoichiometry of the reaction, 1 mole of PCl₅ produces 1 mole of Cl₂, so 0.4087 mol of PCl₅ will produce 0.4087 mol of Cl₂.
To find the volume of Cl₂ at standard temperature and pressure (STP), we use the ideal gas law. At STP (0°C and 1 atm), 1 mole of gas occupies 22.414 L. Thus:
Volume of Cl₂ = moles × molar volume at STP = 0.4087 mol × 22.414 L/mol = 9.164 L
The volume of chlorine gas produced from the decomposition of 85.10 g of PCl₅ is 9.164 liters at STP, rounded to three significant figures.