/*! 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 134 Which of the following substance... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following substances contains the greatest mass of chlorine? (a) \(5.0 \mathrm{~g} \mathrm{Cl}_{2},\) (b) \(60.0 \mathrm{~g}\mathrm{NaClO}_{3}\) (c) \(0.10 \mathrm{~mol} \mathrm{KCl}\), (d) \(30.0 \mathrm{~g} \mathrm{MgCl}_{2}\), (e) \(0.50 \mathrm{~mol} \mathrm{Cl}_{2}\).

Short Answer

Expert verified
The substance with the greatest mass of chlorine is (e) \(0.50 \mathrm{~mol} \mathrm{Cl}_{2}\) with a mass of \(35.45 \mathrm{~g}\).

Step by step solution

01

Find atomic weights and formula weights

Look up the atomic weights of the elements. Chlorine's atomic weight is approximately \(35.45 \mathrm{~g/mol}\). Then, calculate the molar mass for the compounds \(\mathrm{NaClO}_{3}, \mathrm{KCl}, \) and \(\mathrm{MgCl}_{2}\) , which are approximately \(106.44 \mathrm{~g/mol}\), \(74.55 \mathrm{~g/mol}\) and \(95.21 \mathrm{~g/mol}\) respectively.
02

Calculate the mass of Chlorine in each substance

Calculate the mass of chlorine in each substance. For (a), the mass is directly given as \(5.0 \mathrm{~g}\). For (b), the mass of chlorine is \(\frac{1}{3} \times 60.0 \mathrm{~g} = 20.0 \mathrm{~g}\), since \(\mathrm{NaClO}_{3}\) contains one chlorine atom per every three atomic units. For (c), the mass of chlorine is \(0.10 \mathrm{~mol} \times 35.45 \mathrm{~g/mol} = 3.545 \mathrm{~g}\). For (d), the mass of chlorine is \(\frac{2}{3} \times 30.0 \mathrm{~g} = 20.0 \mathrm{~g}\), since \(\mathrm{MgCl}_{2}\) contains two chlorine atoms per every three atomic units. For (e), the mass of chlorine is \(0.50 \mathrm{~mol} \times 70.90 \mathrm{~g/mol} = 35.45 \mathrm{~g}\), since molecular chlorine contains two chlorine atoms.
03

Compare the masses

Compare the masses calculated in step 2. The substance with the greatest mass of chlorine is (e) \(0.50 \mathrm{~mol} \mathrm{Cl}_{2}\) with a mass of \(35.45 \mathrm{~g}\), followed by (a), (b) and (d) with masses of \(20.0 \mathrm{~g}\). The substance with the least mass of chlorine is (c) \(0.10 \mathrm{~mol} \mathrm{KCl}\) with a mass of \(3.545 \mathrm{~g}\).

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.

Atomic Weight
When trying to understand chemical compounds and their reactions, it's important to know about atomic weight. Atomic weight is like the average weight of an atom of an element, based on how it is naturally found with its isotopes. For chlorine (Cl), its atomic weight is about 35.45 g/mol. This value comes from the masses of its two main isotopes, chlorine-35 and chlorine-37.

Atomic weight helps us know more about how much of each element is in a compound. For students, it's key to grasp this concept as it lays the groundwork for calculating other properties, like molar mass.
Molar Mass
Molar mass is the mass of one mole of a given substance. To find it, we simply add up the atomic weights of all the atoms in a molecule of the compound. This is where atomic weights come into play, providing the numbers we need for this calculation.

Take the compound ext{NaClO}_3 as an example. By adding the atomic weights of sodium (Na), chlorine (Cl), and three oxygen (O) atoms, we get a molar mass of 106.44 g/mol.
  • For ext{KCl} , it's calculated as 74.55 g/mol.
  • For ext{MgCl}_2 , we calculate it as 95.21 g/mol.

Understanding molar mass allows you to convert between the mass of a substance and the amount in moles, a key step in chemical equations and reactions.
Chlorine-Containing Compounds
Chlorine plays a key role in a variety of chemical compounds. These are often referred to as "chlorine-containing compounds." In the context of the exercise, we deal with different such compounds like ext{NaClO}_3 , ext{KCl} , ext{MgCl}_2 , and ext{Cl}_2 . Each of these compounds contains chlorine atoms, but the number and arrangement differ.

ext{Cl}_2 is simply two chlorine atoms bonded together. Meanwhile, ext{NaClO}_3 contains one chlorine molecule in a more complex structure combined with sodium and oxygen. Understanding the composition of these compounds helps to analyze their chemical behavior and how much chlorine is in them.
Stoichiometry
Stoichiometry is the part of chemistry that helps us understand the relationship between reactants and products in a chemical reaction. It uses numbers and formulas to predict how much of each substance is needed or produced.

In the exercise, stoichiometry helps us calculate how much chlorine is present in a specified mass or moles of its compounds. For example, to find out how much chlorine is in 0.50 moles of ext{Cl}_2 , we multiply the moles by the molar mass (70.90 g/mol): (0.50 ext{ moles} imes 70.90 ext{ g/mol} = 35.45 ext{ g}) . This calculation method can help solve various real-world chemistry problems, from cooking to industrial manufacturing.

Thus, stoichiometry offers a structured way to think about quantities and how they interact in the world of chemistry.

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

limestone \(\left(\mathrm{CaCO}_{3}\right)\) is decomposed by heating to quicklime \((\mathrm{CaO})\) and carbon dioxide. Calculate how many grams of quicklime can be produced from \(1.0 \mathrm{~kg}\) of limestone.

Hemoglobin \(\left(\mathrm{C}_{2952} \mathrm{H}_{4664} \mathrm{~N}_{812} \mathrm{O}_{832} \mathrm{~S}_{8} \mathrm{Fe}_{4}\right)\) is the oxy- gen carrier in blood. (a) Calculate its molar mass. (b) An average adult has about \(5.0 \mathrm{~L}\) of blood. Every milliliter of blood has approximately \(5.0 \times 10^{9}\) erythrocytes, or red blood cells, and every red blood cell has about \(2.8 \times 10^{8}\) hemoglobin molecules. Calculate the mass of hemoglobin molecules in grams in an average adult.

An impure sample of zinc (Zn) is treated with an excess of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) to form zinc sulfate \(\left(\mathrm{ZnSO}_{4}\right)\) and molecular hydrogen \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) If \(0.0764 \mathrm{~g}\) of \(\mathrm{H}_{2}\) is obtained from \(3.86 \mathrm{~g}\) of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in (b)?

Define limiting reactant and excess reactant. What is the significance of the limiting reactant in predicting the amount of the product obtained in a reaction? Can there be a limiting reactant if only one reactant is present?

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by$$4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

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.