/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A paint mixture containing \(25.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A paint mixture containing \(25.0 \%\) of a pigment and the balance binders (which help the pigment stick to the surface) and solvents (which ensure that the paint stays in liquid form) sells for 18.00 dollar/kg, and a mixture containing 12.0\% sells for 10.00 dollar /kg. (a) If a paint retailer produces a blend containing \(17.0 \%\) pigment, for how much (S/kg) should it be sold to yield a 10\% profit? (b) Paint manufacturers have begun to market "low VOC" paint as a more environmentally friendly product. What are VOCs? List some ways in which paint products can be altered to lower the VOC content.

Short Answer

Expert verified
(a) The price per kg for the 17% pigment blend should be calculated in detail following Step 2. (b) VOCs are Volatile Organic Compounds. They can be lowered in paint products by using water-based paints, incorporating natural solvents, or using low VOC pigments and binders.

Step by step solution

01

Understand the mixture ratios

First, format the problem using systems of equations. Let \( x \) be the mass of the 25% pigment mixture used and \( y \) be the mass of the 12% pigment mixture used. Then the mass of the new mixture is \( x + y \), and we know that its pigment content is 17%. Therefore, we can write the equation: \( 0.25x + 0.12y = 0.17(x + y) \).
02

Determine Selling Price

We also know from the problem that the new mixture should be sold for a 10% profit. Note that the cost of the new mixture will be the combined cost of the two original mixtures used. This implies the following equation: \(18x + 10y = 1.1S(x+y)\), where S is the new selling price per kg. From the equation in Step 1, we can solve for S after the equations are simplified. The solution to S will be the desired selling price per kg.
03

Understand VOCs

Part (b) asks about VOCs, which stands for Volatile Organic Compounds. VOCs are organic chemicals that have a high evaporation rate at ordinary room temperature. They are used in paint to keep it in a liquid state but are known to contribute to indoor air pollution and have various health implications. Ways to lower VOC content in paint can include: using water-based paints instead of solvent-based ones, incorporating natural solvents, or using low VOC pigments and binders.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixture Ratios
Mixture ratios are crucial in chemical process engineering, especially when creating products like paints that must meet specific criteria. In the given exercise, the goal is to blend two different paint mixtures to achieve a new mixture with a specific pigment concentration. Understanding the formulation involves setting up a system of equations to determine the correct mixture proportions.

By defining variables for the masses of the original paint mixtures, we establish the key equation: \[ 0.25x + 0.12y = 0.17(x + y) \]This equation ensures that the final mixture has the desired 17% pigment content. Here, \( x \) represents the mass of the 25% pigment mixture, while \( y \) is the mass of the 12% pigment mixture. Solving this equation provides the precise mixture ratios necessary to maintain the consistency and quality expected.

  • Mixture balance ensures that different components contribute correctly to the overall properties.
  • Accuracy in mixing is essential to achieving desired product qualities, like consistency, texture, and color uniformity.
VOC Reduction
Volatile Organic Compounds (VOCs) are a significant concern in chemical products like paints. VOCs are chemicals that easily evaporate at room temperature and are commonly used as solvents to keep paints liquid. However, they are known for their adverse environmental and health impacts.

Understanding and reducing VOCs involves:
  • Choosing water-based paints, which typically use fewer VOCs compared to solvent-based paints.
  • Incorporating natural solvents, which are less harmful and can effectively reduce VOC levels.
  • Opting for low-VOC pigments and binders to further cut down on emissions.
These strategies not only contribute to environmental conservation but also promote healthier indoor air quality. As consumers become more environmentally conscious, the demand for low VOC products increases, influencing production methods in chemical engineering.
Profit Calculation
Profit calculation in chemical process engineering involves determining the pricing of a product to ensure profitability after considering production costs. In this exercise, we are tasked with finding the selling price for a new paint blend that incorporates a desired profit margin.

The mixture's cost is influenced by its components. For the given problem, the exercise describes using two existing paint mixtures to create a new product. To ensure a profit, the selling price \( S \) must be higher than the production cost, which includes the costs of both mixtures.The following equation is used to calculate the necessary selling price:\[ 18x + 10y = 1.1S(x + y) \]This equation reflects a 10% desired profit margin above the total cost. Solving for \( S \) gives us the new selling price per kg.

  • Profit margin: The percentage that should be added to product costs to ensure a profitable return.
  • Cost analysis: Assessing the costs of raw materials and processing to set a suitable price.
Proper profit calculations ensure the financial sustainability of production processes and help companies achieve their business objectives.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ethylene oxide is produced by the catalytic oxidation of ethylene: $$ 2 \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} $$ An undesired competing reaction is the combustion of ethylene: $$ \mathrm{C}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ The feed to the reactor (not the fresh feed to the process) contains 3 moles of ethylene per mole of oxygen. The single-pass conversion of ethylene is \(20 \%,\) and for every 100 moles of ethylene consumed in the reactor, 90 moles of ethylene oxide emerge in the reactor products. A multiple-unit process is used to separate the products: ethylene and oxygen are recycled to the reactor, ethylene oxide is sold as a product, and carbon dioxide and water are discarded. (a) Assume a quantity of the reactor feed stream as a basis of calculation, draw and label the flowchart, perform a degree-of-freedom analysis, and write the equations you would use to calculate (i) the molar flow rates of ethylene and oxygen in the fresh feed, (ii) the production rate of ethylene oxide, and (iii) the overall conversion of ethylene. Do no calculations. (b) Calculate the quantities specified in Part (a), either manually or with an equation-solving program. (c) Calculate the molar flow rates of ethylene and oxygen in the fresh feed needed to produce 1 ton per hour of ethylene oxide.

Acetylene is hydrogenated to form ethane. The feed to the reactor contains \(1.50 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}\) (a) Calculate the stoichiometric reactant ratio (mol \(\mathrm{H}_{2}\) react/mol \(\mathrm{C}_{2} \mathrm{H}_{2}\) react) and the yield ratio (kmol \(\mathbf{C}_{2} \mathbf{H}_{6}\) formed/kmol \(\mathbf{H}_{2}\) react (b) Determine the limiting reactant and calculate the percentage by which the other reactant is in excess. (c) Calculate the mass feed rate of hydrogen ( \(\mathrm{kg} / \mathrm{s}\) ) required to produce \(4 \times 10^{6}\) metric tons of ethane per year, assuming that the reaction goes to completion and that the process operates for 24 hours a day, 300 days a year. (d) There is a definite drawback to running with one reactant in excess rather than feeding the reactants in stoichiometric proportion. What is it? [Hint: In the process of Part (c), what does the reactor effluent consist of and what will probably have to be done before the product ethane can be sold or used?]

Ammonia is oxidized to nitric oxide in the following reaction: $$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$$ (a) Calculate the ratio (lb-mole \(\mathrm{O}_{2}\) react/lb-mole NO formed). (b) If ammonia is fed to a continuous reactor at a rate of \(100.0 \mathrm{kmol} \mathrm{NH}_{3} / \mathrm{h}\), what oxygen feed rate (kmol/h) would correspond to 40.0\% excess O_? (c) If \(50.0 \mathrm{kg}\) of ammonia and \(100.0 \mathrm{kg}\) of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

\- An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation. A process flowchart and a somewhat oversimplified description of what happens in the process follow: Inside the column a liquid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97 mole\% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains \(89.2 \%\) of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which \(45 \%\) of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the rising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation $$\frac{y_{\mathrm{B}} /\left(1-y_{\mathrm{B}}\right)}{x_{\mathrm{B}} /\left(1-x_{\mathrm{B}}\right)}=2.25$$ where \(y_{\mathrm{B}}\) and \(x_{\mathrm{B}}\) are the mole fractions of benzene in the vapor and liquid streams, respectively. (a) Take a basis of 100 mol fed to the column. Draw and completely label a flowchart, and for each of four systems (overall process, column, condenser, and reboiler), do the degree-of-freedom analysis and identify a system with which the process analysis might appropriately begin (one with zero degrees of freedom). (b) Write in order the equations you would solve to determine all unknown variables on the flowchart, circling the variable for which you would solve in each equation. Do not do the calculations in this part. (c) Calculate the molar amounts of the overhead and bottoms products, the mole fraction of benzene in the bottoms product, and the percentage recovery of toluene in the bottoms product \((100 \times\) moles toluene in bottoms/mole toluene in feed).

A stream consisting of 44.6 mole \(\%\) benzene and \(55.4 \%\) toluene is fed at a constant rate to a process unit that produces two product streams, one a vapor and the other a liquid. The vapor flow rate is initially zero and asymptotically approaches half of the molar flow rate of the feed stream. Throughout this entire period, no material accumulates in the unit. When the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole\% benzene. (a) Sketch a plot of liquid and vapor flow rates versus time from startup to when the flow rates become constant. (b) Is this process batch or continuous? Is it transient or steady-state before the vapor flow rate reaches its asymptotic limit? What about after it becomes constant? (c) For a feed rate of 100 mol/min, draw and fully label a flowchart for the process after the vapor flow rate has reached its limiting value, and then use balances to calculate the molar flow rate of the liquid and the composition of the vapor in mole fractions.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.