/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 27

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 91影视!

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

A catalytic reactor is used to produce formaldehyde from methanol in the reaction $$\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{HCHO}+\mathrm{H}_{2}$$ A single-pass conversion of \(60.0 \%\) is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple-unit process. The production rate of formaldehyde is 900.0 kg/h. (a) Calculate the required feed rate of methanol to the process ( \(\mathrm{kmol} / \mathrm{h}\) ) if there is no recycle. (b) Suppose the unreacted methanol is recovered and recycled to the reactor and the single-pass conversion remains 60\%. Without doing any calculations, prove that you have enough information to determine the required fresh feed rate of methanol (kmol/h) and the rates (kmol/h) at which methanol enters and leaves the reactor. Then perform the calculations. (c) The single-pass conversion in the reactor, \(X_{\mathrm{sp}},\) affects the costs of the reactor \(\left(C_{\mathrm{r}}\right)\) and the separation process and recycle line \(\left(C_{\mathrm{s}}\right) .\) What effect would you expect an increased \(X_{\mathrm{sp}}\) would have on each of these costs for a fixed formaldehyde production rate? (Hint: To get a \(100 \%\) singlepass conversion you would need an infinitely large reactor, and lowering the single-pass conversion leads to a need to process greater amounts of fluid through both process units and the recycle line.) What would you expect a plot of \(\left(C_{\mathrm{r}}+C_{\mathrm{s}}\right)\) versus \(X_{\mathrm{sp}}\) to look like? What does the design specification \(X_{\mathrm{sp}}=60 \%\) probably represent?

Titanium dioxide \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) is used extensively as a white pigment. It is produced from an ore that contains ilmenite \(\left(\mathrm{FeTiO}_{3}\right)\) and ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) .\) The ore is digested with an aqueous sulfuric acid solution to produce an aqueous solution of titanyl sulfate \(\left[(\mathrm{TiO}) \mathrm{SO}_{4}\right]\) and ferrous sulfate (FeSO \(_{4}\) ). Water is added to hydrolyze the titanyl sulfate to \(\mathrm{H}_{2} \mathrm{TiO}_{3},\) which precipitates, and \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) The precipitate is then roasted, driving off water and leaving a residue of pure titanium dioxide. (Several steps to remove iron from the intermediate solutions as iron sulfate have been omitted from this description.) Suppose an ore containing \(24.3 \%\) Ti by mass is digested with an \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) solution, supplied in \(50 \%\) excess of the amount needed to convert all the ilmenite to titanyl sulfate and all the ferric oxide to ferric sulfate \(\left[\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\right] .\) Further suppose that \(89 \%\) of the ilmenite actually decomposes. Calculate the masses (kg) of ore and 80\% sulfuric acid solution that must be fed to produce \(1000 \mathrm{kg}\) of pure \(\mathrm{TiO}_{2}\)

In the production of soybean oil, dried and flaked soybeans are brought into contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil. (a) Draw a flowchart of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract). (b) The soybeans contain 18.5 wt\% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to \(2.0 \mathrm{kg}\) hexane per \(\mathrm{kg}\) beans. The residual solids leaving the extraction unit contain 35.0 wt\% hexane, all of the non-oil solids that entered with the beans, and \(1.0 \%\) of the oil that entered with the beans. For a feed rate of \(1000 \mathrm{kg} / \mathrm{h}\) of dried flaked soybeans, calculate the mass flow rates of the extract and residual solids, and the composition of the extract. (c) The product soybean oil must now be separated from the extract. Sketch a flowchart with two units, the extraction unit from Parts (a) and (b) and the unit separating soybean oil from hexane. Propose a use for the recovered hexane.

A mixture of 75 mole \(\%\) methane and 25 mole \(\%\) hydrogen is burned with \(25 \%\) excess air. Fractional conversions of \(90 \%\) of the methane and \(85 \%\) of the hydrogen are achieved; of the methane that reacts, \(95 \%\) reacts to form \(\mathrm{CO}_{2}\) and the balance reacts to form CO. The hot combustion product gas passes through a boiler in which heat transferred from the gas converts boiler feedwater into steam. (a) Calculate the concentration of \(\mathrm{CO}\) (ppm) in the stack gas. (b) The CO in the stack gas is a pollutant. Its concentration can be decreased by increasing the percent excess air fed to the furnace. Think of at least two costs of doing so. (Hint: The heat released by the combustion goes into heating the combustion products; the higher the combustion product temperature, the more steam is produced.)

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water: $$\begin{aligned} \mathrm{CH}_{4}+\mathrm{O}_{2} & \rightarrow \mathrm{HCHO}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{CH}_{4}+2 \mathrm{O}_{2} & \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ The feed to the reactor contains equimolar amounts of methane and oxygen. Assume a basis of \(100 \mathrm{mol}\) feed/s. (a) Draw and label a flowchart. Use a degree-of-freedom analysis based on extents of reaction to determine how many process variable values must be specified for the remaining variable values to be calculated. (b) Use Equation 4.6-7 to derive expressions for the product stream component flow rates in terms of the two extents of reaction, \(\xi_{1}\) and \(\xi_{2}\) (c) The fractional conversion of methane is 0.900 and the fractional yield of formaldehyde is 0.855 . Calculate the molar composition of the reactor output stream and the selectivity of formaldehyde production relative to carbon dioxide production. (d) A classmate of yours makes the following observation: "If you add the stoichiometric equations for the two reactions, you get the balanced equation $$2 \mathrm{CH}_{4}+3 \mathrm{O}_{2} \rightarrow \mathrm{HCHO}+\mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$$ The reactor output must therefore contain one mole of \(\mathrm{CO}_{2}\) for every mole of HCHO, so the selectivity of formaldehyde to carbon dioxide must be \(1.0 .\) Doing it the way the book said to do it, \(I\) got a different selectivity. Which way is right, and why is the other way wrong?" What is your response?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation
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.