All categories

3 years ago

If 21.42g of KMnO4 is actually produced what is the percent yield

Chemistry

1 answer:

mrs_skeptik [129]3 years ago

4 0

Answer:

Percentage yield = 6.776%

Explanation:

Data Given:

Actual yield of KMnO₄ = 21.42g

Percentage Yield = ?

Formula Used to find Percent yield

Percentage yield = Actual yield/ theoretical yield x 100 (1)

For this Pupose First step is

We have to know the theoretical yield KMnO₄

Potasium permagnate form from MnO₄ and KOH in the presence of Oxygen by heating, in 1st step in second 2K₂MnO₄ react with HCl and give KMnO₄ .

The Reactions of formation of KMnO₄

1st Step

2MnO₄ + 4KOH + O₂ ------------> 2K₂MnO₄ + 2H₂O

2nd Step

2K₂MnO₄ + 4HCl ------------->2 KMnO₄ + H₂O + 4KCl

So form the above equation we come to know it produced 2 Mole of KMnO₄

Now we will calculate the mass of KMnO₄ by mass formulae

mass of KMnO₄ = Number of mole of KMnO₄ x Molar Mass of KMnO₄

Molar Mass of KMnO₄ = 158.034g /mol

Put value in Mass Formula

mass of KMnO₄ = 2 mol x 158.034g/mol

mass of KMnO₄ = 316.1 g

So the theoratical yield per standard reaction = 316.1 g

Now put all values in equation 1

Percentage yield = Actual yield/ theoretical yield x 100

Percentage yield = 21.42g / 316.1 g x 100

Percentage yield = 0.0678 x 100

Percentage yield = 6.776%

You might be interested in

Some molecules in air contain . Explain how atoms from the air can become part of human cells

Tcecarenko [31]

The atoms in air are combined with oxygen present in the air and when we inhale the air, oxygen is moved into the body then atoms in oxygen are carried by red blood cells the blood is pumped to the lungs and when oxygen is transported to the body organ, the atoms in lungs become the part of human cells.

5 0

3 years ago

A sodium bromide solution is added to a beaker containing aqueous chlorine. What would happen?

Darya [45]

Answer:

See detailed explanation.

Explanation:

Hello!

In this case, for the described chemical reaction, we can proceed as follows:

A) For the complete chemical reaction we note down every reacted and produced species as well as the proper balancing process:

$2NaBr(aq)+Cl_2(aq)\rightarrow 2NaCl(aq)+Br_2(g)$

In which gaseous bromine may give off.

B) The dissociated ionic equation requires the ionization of the aqueous species in ions, expect for chlorine which is not ionized:

$2Na^+(aq)+2Br^-(aq)+Cl_2(aq)\rightarrow 2Na^+(aq)+2Cl^-(aq)+Br_2(g)$

C) For the net ionic equation we cancel out the sodium ions as they are at both reactants and products:

$2Br^-(aq)+Cl_2(aq)\rightarrow +2Cl^-(aq)+Br_2(g)$

D) Based on C) we infer that the spectator ions here are the sodium ions.

Best regards!

6 0

3 years ago

Scientific models can never be changed. Please select the best answer from the choices provided T F

docker41 [41]

False, because new information is being discovered all the time so they add the new information to the models to make them accurate.

7 0

3 years ago

Read 2 more answers

El plomo y el yodo forman dos compuestos. En uno, el porcentaje de en masa de plomo es del 44,94% y en el otro es de 62,02%. Cal

nikdorinn [45]

Answer:

0.8162 gramos de plomo por gramo de yodo

1.633 gramos de plomo por gramo de yodo

Explanation:

Asumiendo una base de 100 gramos para cada compuesto:

Primer compuesto:

Gramos plomo: 44.94g

Gramos de yodo: 100-44.94g = 55.06g

Así, la masa de plomo por gramos de yodo para el primer compuesto es:

44.94g plomo / 55.06g Yodo =

0.8162 gramos de plomo por gramo de yodo

Segundo compuesto:

Gramos plomo: 62.02g

Gramos de yodo: 100-62.02g = 37.98g

La masa de plomo por gramos de yodo para el segundo compuesto es:

62.02g plomo / 37.98g Yodo =

1.633 gramos de plomo por gramo de yodo

8 0

3 years ago

What is the pressure, in atmospheres, of a 0.108-mol sample of helium gas at a temperature of 20.0°C if its volume is 0.505 L?

SpyIntel [72]

Answer:

P = 5.14ATM

Explanation:

Number of moles = 0.108moles

Temperature (T) = 20°C = 20 + 273.15 = 293.15K

Volume V = 0.505L.

Pressure (P) = ?

R = 0.082J/mol.K

From ideal gas equation,

PV = nRT

P = nRT / V

P = (0.108 * 0.082 * 293.15) / 0.505

P = 2.596 / 0.505

P = 5.14ATM

The pressure of the gas is 5.14ATM

5 0

3 years ago