Question

A 2.00-l reaction vessel, initially at 298 k, contains chlorine gas at a partial pressure of 337 mmhg and fluorine gas at a partial pressure of 729 mmhg. identify the limiting reactant and determine the theoretical yield of clf3 in grams.

Answer 1

It is 189 k of clf3 ok ok ok ok ok ok ok ok ok ok ok

Answer 2

Answer: The theoretical yield of
`ClF_3` is 2.24 g

Step-by-step explanation:

The equation given by ideal gas follows:

`PV=nRT` .......(1)

where, P = pressure of the gas

V = Volume of the gas

T = Temperature of the gas

R = Gas constant =
`62.3637\text{ L mmHg }mol^(-1)K^(-1)`

n = number of moles of gas

• For fluorine gas:

We are given:

`P=337mmHg\\V=2.00L\\T=298K\\n=?`

Putting values in equation 1, we get:

`337mmHg* 2.00L=n* 62.3637\text{ L mmHg }mol^(-1)K^(-1)* 298K\\\\n=(337* 2.00)/(62.3637* 298)=0.0363mol`

• For chlorine gas:

We are given:

`P=729mmHg\\V=2.00L\\T=298K\\n=?`

Putting values in equation 1, we get:

`729mmHg* 2.00L=n* 62.3637\text{ L mmHg }mol^(-1)K^(-1)* 298K\\\\n=(729* 2.00)/(62.3637* 298)=0.0784mol`

The chemical equation for the reaction of chlorine gas and fluorine gas follows:

`Cl_2+3F_2\rightarrow 2ClF_3`

By Stoichiometry of the reaction:

3 moles of fluorine gas reacts with 1 mole of chlorine gas.

So, 0.0363 moles of fluorine gas will react with =
`(1)/(3)* 0.0363=0.0121mol` of chlorine gas

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

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

By Stoichiometry of the reaction:

3 moles of fluorine gas produces 2 moles of
`ClF_3`

So, 0.0363 moles of fluorine gas will produce =
`(2)/(3)* 0.0363=0.0242mol` of
`ClF_3`

• To calculate the theoretical yield of
  `ClF_3` we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of
`ClF_3` = 92.45 g/mol

Moles of
`ClF_3` = 0.0242 moles

Putting values in above equation, we get:

`0.0242mol=\frac{\text{Mass of }ClF_3}{92.45g/mol}\\\\\text{Mass of }ClF_3=(0.0242mol* 92.45g/mol)=2.24g`

Hence, the theoretical yield of
`ClF_3` is 2.24 g