Gravimetric Analysis Problem

Category: Chemical Engineering Math, Algebra

"Published in Suisun City, California, USA"

A mixture of CuSO₄·5H₂O and FeSO₄·7H₂O weighing 100 g is heated in an oven at 373K (100°C) for the removal of water by hydration (by evaporation). The weight of the mixture after drying (after the removal of water by hydration) is found to be 60 g. Calculate the ratio of CuSO₄·5H₂O to
FeSO₄·7H₂O in the original mixture. (Atomic Weights: Cu = 63, S = 32, O = 16, Fe = 56, and H = 1)

Solution:

The given word problem is about the decomposition of the mixture of CuSO₄·5H₂O and FeSO₄·7H₂O by heating in an oven to remove water molecules which involves the principles of Stoichiometry. The total amount of the original mixture and the total amount of the mixture after hydration are given in the problem. 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. Let's analyze the given word problem as follows

Basis: 100 g of the original mixture
            60 g of the mixture after hydration
            
Let x = be the amount of CuSO₄·5H₂O
      y = be the amount of FeSO₄·7H₂O

The total amount of a mixture before drying is

Molecular Weight Data:

CuSO₄= 159
CuSO₄·5H₂O =  249
FeSO₄= 152
FeSO₄·7H₂O = 278 

Consider the decomposition of CuSO₄·5H₂O as follows:

The weight of CuSO₄ will be equal to

Consider the decomposition of FeSO₄·7H₂O as follows:

The weight of FeSO₄ will be equal to

The total amount of mixture after drying is

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

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

Therefore, the ratio of CuSO₄·5H₂O to FeSO₄·7H₂O in the original mixture is 

Stoichiometry Problem - Material Balance, 17

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

In a textile industry, it is desired to prepare 24% caustic soda solution by weight. Because of very high heat of dissolution of caustic soda in water, the above solution is prepared by two-step process. First, caustic soda is dissolved in the correct quantity of water in a dissolution tank to prepare 50% (by weight) solution. After dissolution and cooling is complete, this solution is taken to a dilution tank where some more water is added for producing 24% by weight caustic soda solution. Assuming no evaporation loss of water in dissolution tank, calculate the weight ratio of water fed to the dissolution tank to bypass water to the dilution tank. Also, calculate the total weight of water used in the process.

Photo by Math Principles in Everyday Life

Solution:

The given word problem is about the preparation of caustic soda solution in two steps which involves the principles of Stoichiometry. The solid NaOH is mixed with some water to make a caustic soda solution and the rest of water is bypassed to dilute the caustic soda solution into a lower concentration. 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 label further the flow diagram as follows

Photo by Math Principles in Everyday Life

Basis: 100 kg of Solid NaOH

Let w = be the total amount of water used in the process
      x = be the amount of water fed to Dissolution Tank
      y = be the amount of water bypassed and fed to Dilution Tank
      z = be the amount of 50% NaOH Solution
      p = be the amount of 24% NaOH Solution

Overall Material Balance before Dissolution Tank:

Overall Material Balance for Dissolution Tank:

Material Balance of NaOH at Dissolution Tank:

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

Overall Material Balance for Dilution Tank:

Substitute the value of z to the above equation, we have

Material Balance of NaOH at Dilution Tank:

Substitute the value of z to the above equation, we have

The value of y will be equal to 

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

Therefore, 

The total weight of water used in the process is 316.667 kg.