Stoichiometry Problem - Material Balance

Category: Chemical Engineering Math, Algebra

"Published in Suisun City, California, USA"

The waste acid from a nitrating process containing 20% HNO₃, 55% H₂SO₄, and 25% H₂0 by weight is to be concentrated by addition of concentrated sulfuric acid containing 95% H₂SO₄ and concentrated nitric acid containing 90% HNO₃ to get desired mixed acid containing 26% HNO₃ and 60% H₂SO₄. Calculate the quantities of waste and concentrated acids required for 1000 kg of desired mixed acid. 

Solution:  

The given word problem is about the mixing of different acids 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 sulfuric acid
      z = be the amount of concentrated nitric acid

Overall Material Balance around the Mixer:

Material Balance of Sulfuric Acid:

Material Balance for Nitric Acid:

Substitute the value 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 = 401.6 kg
Amount of Concentrated Sulfuric Acid = 398.65 kg
Amount of Concentrated Nitric Acid = 199.75 kg 

Rate, Distance, Time - Problem, 5

Category: Algebra

"Published in Newark, California, USA"

From Manila to a certain town in Batangas is some 224 miles. How long does it ordinarily take to reach the town, if by traveling 8 miles per hour faster, the trip is reduced by half an hour?

Solution:

The given word problem above is about rate, distance, and time problem with some conditions. Lets analyze the given word problem as follows:

Let x = be the rate in miles per hour
      y = be the travel time in hour

We know that 

  
The first working equation will be

If the statement says, "...if by traveling 8 miles per hour faster, the trip is reduced by half an hour?", then the second working equation will be

To eliminate y, substitute the value of y from the first working equation to the second equation, we have

Solve the quadratic equation by completing the square

Take the square root on both sides of the equation

In solving for the rate and time, we have to choose the positive value at the right side of the equation. Hence, the rate is

Therefore, the usual travel time is

 

Permutation Problems, 3

Category: Algebra, Statistics

"Published in Newark, California, USA"

A passenger train consists of 3 baggage cars, 5 day coaches, and 2 parlor cars. In how many ways can the train be arranged if the 3 baggage cars must come up front?

Solution:

The given word problem is about permutation problem because it involves the number of ways in arranging the objects or things. 

This permutation type is different and it is called a Distinguishable Permutation. 

If a set of n objects consists of k different kinds of objects with n₁, objects of the first kind, n₂ objects of the second kind, n₃ objects of the third kind, and so on, where n₁ + n₂ + ......... + n_k = n, then the number of distinguishable permutations of these objects is 

Now, in the given problem, if

n = 10 train cars in total
n₁ = 3 baggage cars
n₂ = 5 day coaches
n₃ = 2 parlor cars

then, the number of ways in arranging the 10 train cars will be equal to

If the 3 baggage cars must come up front, then the number of ways will be equal to

You have to multiply the previous ways by 3! because the three baggage cars themselves can be arranged in different ways at the front. Therefore, the final answer is