Stoichiometry Problem - Material Balance, 7

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

The waste acid from a nitrating process contains 30% H₂SO₄, 35% HNO₃, and 35% H₂O by weight. The acid is to be concentrated to contain 39% H₂SO₄ and 42% HNO₃ by addition of 10% oleum and concentrated nitric acid containing 72% HNO₃ by weight. Calculate the amount of waste acid, oleum, and concentrated nitric acid to be mixed to get 1000 kg of desired mixed acid.

Solution:

The given word problem is about the mixing of acids and oleum which involves the principles of Stoichiometry. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. Since all incoming substances are acids, then there's no chemical reactions involved in the mixture. To illustrate the given problem, it is better to draw the flow diagram as follows

Photo by Math Principles in Everyday Life

Basis: 1000 kg of Desired Mixed Acid

Let x = be the amount of Waste Acid
      y = be the amount of Concentrated Nitric Acid
      z = be the amount of Oleum

Overall Material Balance of Mixer:

Material Balance of HNO₃:

Oleum is a chemical substance which consist of the mixture of sulfuric acid and sulfur trioxide. If you have 100 kg of 10% oleum, then you have 10 kg of SO₃ and 90 kg of H₂SO₄. 

We can express SO₃ in terms of H₂SO₄ by gravimetric analysis as follows

From the chemical equation above, if 1 kmol SO₃ = 1 kmol H₂SO₄, then it follows that their molecular weight will be equal to 80 kg SO₃ = 98 kg H₂SO₄.

The weight of H₂SO₄ in SO₃ can be calculated as follows

The total weight of H₂SO₄ in oleum is

Hence, the % of H₂SO₄ in solution is

Oleum is a highly concentrated H₂SO₄ than the regular concentrated H₂SO₄ because the % H₂SO₄ is more than 100%.

 Material Balance of H₂SO₄:

Substitute the values of y and z to the first equation, we have

Substitute the value of x to the second equation, we have

Substitute the value of x to the third equation, we have

Therefore,

Amount of Waste Acid = 159.93 kg
Amount of Concentrated HNO₃ = 505.57 kg
Amount of 10% Oleum = 334.50 kg

Stoichiometry Problem - Material Balance, 6

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

A sample of coal is found to contain 63% carbon and 24% ash on weight basis. The analysis of refuse after combustion shows 7% carbon and the rest ash. Calculate the percentage of the original carbon unburnt in the refuse. 

Solution:

The given word problem is about the combustion of a coal in a furnace which involves the principles of Stoichiometry. When you add the percent compositions of a coal, it will not be equal to 100%. The rest of the composition of a coal is either water or other substances that are mostly combustible. During the combustion of a coal, the components like carbon, water, ash, and other substances are converting into flue gas and refuse. The flue gas is a mixture of carbon dioxide, carbon monoxide, water vapor, oxygen, nitrogen, and other oxide gases. The refuse is mostly ash but sometimes there are unburnt carbon or coal in the refuse. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. To illustrate the problem, it is better to draw the flow diagram as follows

Photo by Math Principles in Everyday Life

Since the amount of coal, flue gas, and refuse are not given in the problem, let's assign a basis of 100 kg of coal in the problem.

Basis: 100 kg of coal

Let x = be the amount of refuse
      y = be the amount of flue gas

Material Balance of Ash:

Therefore,

Stoichiometry Problem - Material Balance, 5

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

An evaporator system concentrating a weak liquor from 5% to 50% solids handles 100 kg of solid per hour. If the same system is to concentrate a weak liquor from 4% to 35%, find the capacity of the system in terms of solids that can be handled per hour assuming water evaporation capacity to be same in both the cases. 

Solution: 

The given word problem is about the evaporation of a liquid from weak to thick liquor which involves the principles of Stoichiometry. There are two evaporators in the given problem with different cases but the water evaporation capacity is the same for both cases. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. To illustrate the problem, it is better to draw the flow diagram as follows

Photo by Math Principles in Everyday Life

Case I:

Basis: 100 kg/hr of solid handling capacity of the evaporator

Let x₁ = be the amount of weak liquor
      y₁ = be the amount of thick liquor
      z₁ = be the amount of water evaporated

Amount of solids in weak liquor:

Amount of solids in thick liquor:

Overall Material Balance of Evaporator:

For Case II, the water evaporation capacity is the same as with Case I. And so, z₁ = z₂ = 1800 kg/hr of water evaporated.  

Photo by Math Principles in Everyday Life

Case II:

Basis: 1800 kg/hr of water evaporated

Let x₂ = be the amount of weak liquor
      y₂ = be the amount of thick liquor
      z₂ = be the amount of water evaporated

Overall Material Balance of Evaporator:

Material Balance of Solids:

 

Substitute the value of x₂ to the first equation, we have

Hence, the amount of weak liquor is

Therefore, the solid handling capacity of the evaporator is

or

Stoichiometry Problem - Material Balance, 4

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

An evaporator is fed with 15,000 kg/hr of a solution containing 10% NaCl, 15% NaOH, and the rest water. In the operation, water is evaporated and NaCl is precipitated as crystals. The thick liquor leaving the evaporator contains 45% NaOH, 2% NaCl, and the rest water. Calculate:

(a) kg/hr of water evaporated,
(b) kg/hr of salt precipitated, and
(c) kg/hr of thick liquor.

Solution:

The given word problem is about the evaporation of caustic liquid into water vapor, salt precipitate, and thick liquor which involves the principles of Stoichiometry. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. To illustrate the problem, it is better to draw the flow diagram as follows

Photo by Math Principles in Everyday Life

Basis: 15,000 kg/hr of weak liquor


Let x = be the amount of water evaporated
      y = be the amount of thick liquor
      z = be the amount of salt precipitated


Overall Material Balance of Evaporator:

Material Balance of NaOH:

Material Balance of NaCl:

Substitute the values of y and z to the first equation which is the Overall Material Balance of Evaporator, we have

Therefore,


Amount of Water Evaporated = 8,600 kg/hr
Amount of Salt Precipitated = 1,400 kg/hr
Amount of Thick Liquor = 5,000 kg/hr