Question

A 6.00 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?

Answer 1

Answer: The final pressure in the vessel will be 0.965 atm

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For phosphorus trichloride:

Given mass of phosphorus trichloride = 20.0 g

Molar mass of phosphorus trichloride = 137.3 g/mol

Putting values in equation 1, we get:

`\text{Moles of phosphorus trichloride}=(20.0g)/(137.3g/mol)=0.146mol`

• For oxygen gas:

Given mass of oxygen gas = 3.15 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(3.15g)/(32g/mol)=0.098mol`

The chemical equation for the reaction of phosphorus trichloride and oxygen gas follows:

`2PCl_3+O_2\rightarrow 2POCl_3`

By Stoichiometry of the reaction:

2 moles of phosphorus trichloride reacts with 1 mole of oxygen gas

So, 0.146 moles of phosphorus trichloride will react with =
`(1)/(2)* 0.146=0.073mol` of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, phosphorus trichloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of phosphorus trichloride produces 2 moles of
`POCl_3`

So, 0.146 moles of phosphorus trichloride will produce =
`(2)/(2)* 0.146=0.146mol` of
`POCl_3`

To calculate the pressure of the vessel, we use the equation given by ideal gas follows:

`PV=nRT`

where,

P = pressure of the vessel = ?

V = Volume of the vessel = 6.00 L

T = Temperature of the vessel =
`210^oC=[210+273]K=483K`

R = Gas constant =
`0.0821\text{ L. atm }mol^(-1)K^(-1)`

n = number of moles = 0.146 moles

Putting values in above equation, we get:

`P* 6.00L=0.146mol* 0.0821\text{ L atm }mol^(-1)K^(-1)* 483K\\\\P=(0.146* 0.0821* 483)/(6.00)=0.965atm`

Hence, the final pressure in the vessel will be 0.965 atm

Answer 2

The final pressure in the vessel is 1.13 atm

Step-by-step 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