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mamaluj [8]
3 years ago
10

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?
Chemistry
1 answer:
Ronch [10]3 years ago
3 0

<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.

<u>Explanation:</u>

  • <u>For potassium permanganate:</u>

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

  • <u>For hydrochloric acid:</u>

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.

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As we have the mass of the two reactants, we must determine the limiting reactant.

Let's convert to moles, the mass of each reactant. (mass / molar mass)

21.1 g / 36.45 g/mol = 0.579 moles of HCl

46.3 g / 40g/mol = 1.15 moles of NaOH

Ratio is 1:1, so it is obviously that the limiting reactant is the HCl. For 1.15 moles of NaOH, i need the same amount of acid, but I only have 0.579 moles

Let's work with the products now. Ratio is 1:1 again, so If I have 0.579 moles of acid, I can produce 0.579 moles of H₂O.

How many grams are 0.579 moles of water? We should find it out as this

mol . molar mass = mass → 0.579 mol . 18 g/mol = 10.4 g

We were told that the production of water was 9.17 g, so let's determine the percent yield as this:

(Yield produced / Theoretical yield) . 100 =

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Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

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= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

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Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

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if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

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V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

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