Respuesta :
Answer: The volume of chlorine gas produced in the reaction is 2.06 L.
Explanation:
- For potassium permanganate:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of potassium permanganate = 6.23 g
Molar mass of potassium permanganate = 158.034 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol[/tex]
- For hydrochloric acid:
To calculate the moles of hydrochloric acid, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
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:
[tex]6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol[/tex]
- For the reaction of potassium permanganate and hydrochloric acid, the equation follows:
[tex]2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O[/tex]
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 = [tex]\frac{2}{16}\times 0.27=0.033moles[/tex] 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 = [tex]\frac{5}{16}\times 0.27=0.0843moles[/tex] of chlorine gas.
- To calculate the volume of gas, we use the equation given by ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 1.05 atm
V = Volume of gas = ? L
n = Number of moles = 0.0843 mol
R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]40^oC=[40+273]K=313K[/tex]
Putting values in above equation, we get:
[tex]1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L[/tex]
Hence, the volume of chlorine gas produced in the reaction is 2.06 L.
