A6.00 l vessel contains 20.0 g of pcl3 and 3.15 g of o2 at 15.0 ∘c. the vessel is heated to 210 ∘c, and the contents react to give pocl3. what is the final pressure in the vessel, assuming that the reaction goes to completion and that all reactants and products are in the gas phase?

The final pressure in the vessel is 1.13 atm

Explanation:

Step 1: Data given

Volume of the vessel = 6.00 L

Mass of PCl3 = 20.0 grams

Mass of O2 = 3.15 grams

Temperature = 15.0 °C

The vessel is heated to 210°C

Molar mass of PCl3 = 137.33 g/mol

Step 2: The balanced equation

2PCl3 + O2 → 2POCl3

Step 3: Calculate moles PCl3

MolesPCl3 = mass PCl3 / molar mass PCl3

Moles PCl3 = 20.0 grams / 137.33 g/mol

Moles PCl3 =0.146 moles PCl3

Step 4: Calculate moles O2

Moles O2 = 3.15 grams/ 32.0 g/mol

Moles O2 = 0.0984 moles O2

Step 5: Calculate the limiting reactant

PCl3 is the limiting reactant. It will completely be consumed(0.146 moles). So at completion there is no PCl3 remaining.

O2 is in excess. There will react 0.146/2 = 0.073 moles. There will remain 0.0984 - 0.073 = 0.0254 moles O2

Step 6: Calculate moles POCl3

For 2 moles PCl3 we need 1 mol O2 to produce 2 moles POCl3

For 0.146 moles PCl3 we'll have 0.146 moles POCl3

Step 7: Calculate final pressure

p*V = n*R*T

p = (n*R*T)/V

⇒ with n = the number of moles = 0.146 moles of POCl3 produced + 0.0254 moles O2 remaining = 0.1714 moles gas

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 210 +273 = 483 Kelvin

⇒ with V = the volume = 6.00 L

p = (0.1714 *0.08206 * 483) / 6.00

p = 1.13 atm

The final pressure in the vessel is 1.13 atm

answer

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