Question

one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40 C and a pressure of 1.05 atm can be produced by the reaction of 6.23 g KMnO4 with 45.0 ml of 6.00 m HCl?

Answer 1

Answer: The volume of chlorine gas produced in the reaction is 2.06 L.

Step-by-step explanation:

• For potassium permanganate:

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

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

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

`\text{Moles of potassium permanganate}=(6.23g)/(158.034g/mol)=0.039mol`

• For hydrochloric acid:

To calculate the moles of hydrochloric acid, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

`6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol`

• For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

`2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O`

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with =
`(2)/(16)* 0.27=0.033moles` of potassium permanganate.

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

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

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with =
`(5)/(16)* 0.27=0.0843moles` of chlorine gas.

• To calculate the volume of gas, we use the equation given by ideal gas equation:

`PV=nRT`

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant =
`0.0820\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the gas =
`40^oC=[40+273]K=313K`

Putting values in above equation, we get:

`1.05atm* V=0.0843* 0.0820\text{ L atm }mol^(-1)K^(-1)* 313K\\\\V=2.06L`

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.