/*! 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 115 Potash is any potassium mineral ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride ( \(\mathrm{KCl}\) ) and potassium sulfate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)\). Potash production is often reported as the potassium oxide \(\left(\mathrm{K}_{2} \mathrm{O}\right)\) equivalent or the amount of \(\mathrm{K}_{2} \mathrm{O}\) that could be made from a given mineral. (a) If \(\mathrm{KCl}\) costs \(\$ 0.055\) per \(\mathrm{kg}\), for what price (dollar per \(\mathrm{kg})\) must \(\mathrm{K}_{2} \mathrm{SO}_{4}\) be sold in order to supply the same amount of potassium on a per dollar basis? (b) What mass (in kg) of \(\mathrm{K}_{2} \mathrm{O}\) contains the same number of moles of \(\mathrm{K}\) atoms as \(1.00 \mathrm{~kg}\) of \(\mathrm{KCl} ?\)

Short Answer

Expert verified
(a) $0.18 per kg for Kâ‚‚SOâ‚„. (b) 1.263 kg of Kâ‚‚O.

Step by step solution

01

Determine Molar Masses

First, we need to calculate the molar masses of the compounds involved. - Molar mass of KCl = (39.10 for K) + (35.45 for Cl) = 74.55 g/mol. - Molar mass of Kâ‚‚SOâ‚„ = (2*39.10 for K) + (32.07 for S) + (4*16.00 for O) = 174.27 g/mol. - Molar mass of Kâ‚‚O = (2*39.10 for K) + (16.00 for O) = 94.20 g/mol.
02

Calculate Moles of Potassium in 1 kg of KCl and Kâ‚‚SOâ‚„

For KCl: - 1 kg (or 1000 g) of KCl contains 1000 / 74.55 = 13.41 moles of KCl. - Since 1 mole of KCl contains 1 mole of K, there are 13.41 moles of K in 1 kg of KCl. For Kâ‚‚SOâ‚„: - Each mole of Kâ‚‚SOâ‚„ contains 2 moles of K. - Therefore, 1 mole of Kâ‚‚SOâ‚„ contains 2*39.10 = 78.2 g of K per 174.27 g of Kâ‚‚SOâ‚„.
03

Calculate Price per kg for Kâ‚‚SOâ‚„

Given KCl costs $0.055/kg, and it provides 13.41 moles of K per kg. For Kâ‚‚SOâ‚„ to supply the same moles of K, - Cost per mole of K = 0.055 / 13.41 = $0.0041 per mole of K from KCl. - To find the price of Kâ‚‚SOâ‚„, calculate mass equivalent needed to provide same moles: - (1000 g / 174.27 g/mol) * 78.2 g of K per kg of Kâ‚‚SOâ‚„ = 44.92 moles of K per kg of Kâ‚‚SOâ‚„. - Therefore, Kâ‚‚SOâ‚„ to supply 13.41 moles of K costs 13.41 * $0.0041 = $0.055 per kg for the same moles of K.
04

Calculate Mass of Kâ‚‚O Equivalent to 1 kg of KCl

From Step 2, 1 kg of KCl contains 13.41 moles of K. - Each mole corresponds to 94.20 g/mol of Kâ‚‚O, - So, the mass of Kâ‚‚O needed to provide 13.41 moles of K would be 13.41 moles * 94.20 g/mol = 1262.902 g or 1.263 kg of Kâ‚‚O.

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.

Molar Mass Calculation
Calculating the molar mass of a substance is essential for understanding its properties, such as how it behaves in a reaction. Molar mass represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). This calculation involves adding up the atomic masses of all the elements in a compound. For instance, potassium chloride (KCl) has a molar mass calculated as follows: the atomic mass of potassium (K) is 39.10 g/mol and that of chlorine (Cl) is 35.45 g/mol. Combine these, and the molar mass of KCl is 74.55 g/mol.

Similarly, to calculate the molar mass of potassium sulfate (Kâ‚‚SOâ‚„), you need to sum the atomic masses of two potassium atoms, one sulfur atom, and four oxygen atoms. This is done by:
  • 2 x 39.10 g/mol for potassium = 78.20 g/mol
  • 32.07 g/mol for sulfur
  • 4 x 16.00 g/mol for oxygen = 64.00 g/mol
Adding these values gives us a total of 174.27 g/mol for Kâ‚‚SOâ‚„. By understanding how to perform these calculations, you gain a clearer picture of a compound's composition and can predict its behavior in chemical reactions.
Potassium Compounds
Potassium compounds are pivotal in various applications, particularly in agriculture. They can be found in fertilizers, helping to deliver essential nutrients to plants. Potassium chloride (KCl) and potassium sulfate (Kâ‚‚SOâ‚„) are two major sources of potassium used in fertilizers.

KCl is one of the most common forms of potassium applied in agriculture. It is highly soluble in water, making the potassium readily available for plant uptake. However, due to its chloride content, it may not be suitable for chloride-sensitive crops.

On the other hand, Kâ‚‚SOâ‚„ not only provides potassium for plant growth, but also offers sulfur, which is another vital nutrient. Sulfur helps in chlorophyll formation and protein synthesis in plants. This dual-nutrient benefit makes Kâ‚‚SOâ‚„ a valuable option for farmers seeking to enhance plant growth without introducing chlorides.

Though both KCl and Kâ‚‚SOâ‚„ serve similar purposes, their economic value can differ. It is important to evaluate their costs relative to the amount of potassium they supply, as we discussed with molar mass calculations and pricing strategies in fertilizer chemistry.
Fertilizer Chemistry
Fertilizer chemistry is crucial for understanding how different components in fertilizers benefit plant growth. Fertilizers are substances that supply essential nutrients to plants, supporting various physiological processes. In the context of potassium fertilizers, knowing the chemical forms of potassium in fertilizers helps determine their effectiveness.

Potassium is a key macronutrient that plants need to photosynthesize, grow, and develop strong structures. In fertilizers, potassium is often represented as a part of its oxide form, e.g., potassium oxide (Kâ‚‚O). This is because Kâ‚‚O gives us a straightforward measure of available potassium, even if it is not present in this form in the fertilizer directly.

Converting compounds like KCl and Kâ‚‚SOâ‚„ to their Kâ‚‚O equivalent is common practice. This allows for a standardized comparison between different potassium fertilizers. For example, Kâ‚‚SOâ‚„ provides potassium similar to Kâ‚‚O, but with the added advantage of supplying sulfur, as illustrated in pricing strategies based on mole calculations.

The nuanced understanding of fertilizer chemistry supports informed decision-making for optimal crop production. By grasping the concepts of molar mass and the role of different potassium compounds, growers can more effectively choose the right fertilizers for their specific agricultural needs.

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

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, 0.507 mole of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

When potassium cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation: $$ \mathrm{KCN}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{KCl}(a q)+\mathrm{HCN}(g) $$ If a sample of \(0.140 \mathrm{~g}\) of \(\mathrm{KCN}\) is treated with an excess of \(\mathrm{HCl}\), calculate the amount of \(\mathrm{HCN}\) formed, in grams.

Which of the following has more atoms: \(1.10 \mathrm{~g}\) of hydrogen atoms or \(14.7 \mathrm{~g}\) of chromium atoms?

A reaction having a 90 percent yield may be considered a successful experiment. However, in the synthesis of complex molecules such as chlorophyll and many anticancer drugs, a chemist often has to carry out multiple-step synthesis. What is the overall percent yield for such a synthesis, assuming it is a 30 step reaction with a 90 percent yield at each step?

When heated, lithium reacts with nitrogen to form lithium nitride: $$ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s) $$ What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) are heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2} ?\) If the actual yield of \(\mathrm{Li}_{3} \mathrm{~N}\) is \(5.89 \mathrm{~g},\) what is the percent yield of the reaction?

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.