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mixas84 [53]
1 year ago
7

calculate the mass of HNO3 obtained when 20.0g of KNO3 reacts completely with excess concentrated H2SO4

Chemistry
1 answer:
iragen [17]1 year ago
8 0

The mass of HNO3 obtained is 12.46 g from the given reaction.

The chemical reaction is :

Potassium Nitrate + Sulphuric acid ------> Potassium Bisulphate + Nitric acid.

The above reaction is basically used for the making of nitric acid.

Less than 200 degree celcius the Nitric acid if more than 200 degree celcius then it will form Potassium Sulphate.

Now,

1 mol of KNO3 reacts with1 mol of concentrated H2SO4 to give 1 mol HNO3

Molar Mass of KNO3 = 101.1 g

Molar mass of H2SO4 = 98.07 g

Mass of H2SO4 that can react with 20 g of KNO3 = (20 g* 98.07 g )/ 101.1 g

= 19.4 g

Molar mass of nitric acid= 63.01 g

Hence mass of nitric acid  produced by 20 g of KNO3 = (20 g * 63.01 g ) / 101.1 g = 12.46 g

To know more about Nitric acid here :

brainly.com/question/20934865?referrer=searchResults

#SPJ1

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Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar=
IceJOKER [234]

Answer:

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

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<h3>pH = 4.27</h3>

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

<h3>0.03%</h3>
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Enthalpy is a state function

Explanation:

The Hess's law allows us to determine the enthalpy change of a reaction because enthalpy is a state function. It does not depend on the individual path take in going from reactants to products in the reaction.

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