one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40

Question

2 years ago

10

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?

1 answer:

Ronch [10] · Answer 1 · 2023-04-03T00:10:10+03:00

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

Explanation:

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:

$\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol$

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 = $\frac{2}{16}\times 0.27=0.033moles$ 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 = $\frac{5}{16}\times 0.27=0.0843moles$ 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:

$1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L$

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