Stoichiometry Problem - Material Balance, 11

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

2,000 kg of wet solids containing 70% solids by weight are fed to a tray dryer where it is dried by hot air. The product finally obtained is found to contain 1% moisture by weight. Calculate

(a) the weight of water removed from wet solids, and
(b) the weight of product obtained.

Solution:

The given word problem is about drying of a wet solids by hot air dryer 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 given problem, it is better to draw the flow diagram as follows 

Photo by Math Principles in Everyday Life

 Basis: 2,000 kg of solids

Let x = be the weight of Product Solids
      y = be the weight of Water Removed 

Overall Material Balance of Hot Air Dryer:

 
Material Balance of Solids:

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

Therefore,

Weight of Product Solids = 1,414.141 kg
Weight of Water Removed = 585.859 kg

Stoichiometry Problem - Material Balance, 10

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

A mixture of phenol and water forms two separate liquid phases, one rich in phenol and other rich in water, composition of layers is 70% and 9% (by weight) phenol respectively. If 500 kg of phenol and 700 kg of water are mixed and layers allowed to separate, what will be the weights of two layers?

Solution:

The given word problem is about liquid-liquid extraction of phenol-water mixture which involves the principles of Stoichiometry. Liquid–liquid extraction also known as solvent extraction and partitioning, is a method to separate compounds based on their relative solubility in two different immiscible liquids, usually water and an organic solvent. It is an extraction of a substance from one liquid into another liquid phase. Liquid–liquid extraction is a basic technique in chemical laboratories, where it is performed using a separatory funnel.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 given problem, it is better to draw the flow diagram as follows  

Photo by Math Principles in Everyday Life

Basis: 500 kg of Phenol and 700 kg of Water

Let x = be the amount of Phenol Rich Layer
      y = be the amount of Water Rich Layer

Overall Material Balance of Liquid-liquid Extractor:

Material Balance of Phenol:

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

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

Therefore,

Amount of Phenol Rich Layer = 642.419 kg
Amount of Water Rich Layer = 557.581 kg

Stoichiometry Problem - Material Balance, 9

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

10,000 kg/hr of solution containing 20% methanol is continuously fed to a distillation column. Distillate (product) is found to contain 98% methanol and waste solution from the column carries 1% methanol. All percentages are by weight. Calculate

(a) the mass flow rates of distillate and bottom product, and
(b) the percent loss of methanol.

Solution:

The given word problem is about the distillation of methanol solution which involves the principles of Stoichiometry. Distillation is a method of separating mixtures based on differences in volatility of components in a boiling liquid mixture. Distillation is a unit operation, or a physical separation process, and not a chemical reaction.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 given problem, it is better to draw the flow diagram as follows  

Photo by Math Principles in Everyday Life

Basis: 10,000 kg/hr of Feed

Let x = be the mass flow rate of Distillate
      y = be the mass flow rate of Waste Solution

Overall Material Balance of Distillation Column:

Material Balance of Methanol:

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

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

Therefore,

Mass Flow Rate of Distillate =  1,958.763 kg/hr
Mass Flow Rate of Waste Solution = 8,041.237 kg/hr

Next, we can calculate the percent loss for methanol as follows