/*! 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 181 An ionic compound \(\mathrm{MX}_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 181

An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$\mathbf{M}+\mathbf{X}_{2} \longrightarrow \mathbf{M X}_{3}$$ A \(0.105-\mathrm{g}\) sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of \(54.47 \%\) X by mass. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\), and what is the correct name for \(\mathrm{MX}_{3} ?\) Starting with 1.00 g each of \(M\) and \(X_{2}\), what mass of \(M X_{3}\) can be prepared?

Short Answer

Expert verified
We have identified M as Rubidium (Rb) and X as Chlorine (Cl), making the compound Rubidium Chloride (RbCl鈧�). Starting with 1.00 g each of Rb and Cl鈧�, we can prepare 2.20 g of RbCl鈧�.

Step by step solution

01

Calculate the molar mass of X鈧�

To calculate the molar mass of X鈧�, we'll use the mass of X鈧� and the number of molecules available: Molar mass of X鈧� = (Mass of X鈧�) / (Number of molecules of X鈧� / Avogadro's number) Avogadro's number = 6.022 脳 10虏鲁 Number of moles of X鈧� = (8.92 脳 10虏鈦� molecules) / (6.022 脳 10虏鲁) = 0.001482 moles Molar mass of X鈧� = 0.105 g / 0.001482 moles = 70.81 g/mol
02

Calculate the atomic mass of X

Since X鈧� is a diatomic molecule, the atomic mass of X can be calculated by dividing the molar mass of X鈧� by 2: Atomic mass of X = Molar mass of X鈧� / 2 Atomic mass of X = 70.81 g/mol / 2 = 35.40 g/mol
03

Calculate the molar mass of MX鈧� and identify M and X

We know that the mass of X in MX鈧� is 54.47%. Therefore, the mass of M in MX鈧� will be 100% - 54.47% = 45.53%. Let molar mass of MX鈧� = A g/mol Mass of X in 1 mol of MX鈧� = 3 脳 atomic mass of X = 3 脳 35.40 g/mol = 106.20 g/mol Mass of M in 1 mol MX鈧� = mass of M in MX鈧� = A 脳 45.53% Now, mass of X in MX鈧� = A 脳 54.47% 106.20 g/mol = A 脳 54.47% We can solve for A: A = 106.20 g/mol / 54.47% = 194.95 g/mol Now that we have the molar mass of MX鈧�, we can determine the mass of M in 1 mol of MX鈧� and identify M and X: Mass of M = A 脳 45.53% = 194.95 g/mol 脳 45.53% = 88.75 g/mol We can now identify M and X: M 鈮� 89 g/mol (Rb, Rubidium) X 鈮� 35.5 g/mol (Cl, Chlorine) Thus, we have found the elements in MX鈧� to be Rubidium and Chlorine, so the compound is Rubidium Chloride (RbCl鈧�).
04

Calculate the mass of MX鈧� using 1.00g each of M and X鈧�

We start with 1.00 g of M (Rb) and 1.00 g of X鈧� (Cl鈧�). Since M and X鈧� react in a 1:1 ratio, we can find which reactant is the limiting reactant. Number of moles of M = 1.00 g / (atomic mass of Rb) = 1.00 g / 88.75 g/mol = 0.01127 moles Number of moles of X鈧� = 1.00 g / (molar mass of Cl鈧�) = 1.00 g / 70.81 g/mol = 0.01412 moles Since the ratio of M:X鈧� in the reaction is 1:1, Rb is the limiting reactant. Now, for each mole of Rb, one mole of RbCl鈧� is formed. So, 0.01127 moles of Rb reacts completely forming 0.01127 moles of RbCl锛�. Let's calculate the mass of RbCl鈧� formed: Mass of RbCl鈧� = 0.01127 moles 脳 (molar mass of RbCl鈧�) = 0.01127 moles 脳 194.95 g/mol = 2.20 g Hence, we can prepare 2.20 g of RbCl鈧� using 1.00 g each of Rb and Cl鈧�.

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.

Chemical Equations
Chemical equations represent the substances involved in a chemical reaction. They show the reactants being transformed into products, and are crucial for understanding chemical processes. In the case of ionic compounds, such as in our exercise, the equation must balance, meaning the number of each type of atom must be equal on both sides.

For example, in the unbalanced equation given, \( \mathbf{M} + \mathbf{X}_2 \longrightarrow \mathbf{MX}_3 \), each molecule of \( X_2 \) reacts with atoms of \( M \) to form \( MX_3 \).
  • The equation includes diatomic chlorine, \( \mathbf{Cl}_2 \), which naturally exists as pairs of atoms.
  • You need to properly balance the number of molecules and atoms to make sure there is an equal number on each side.
The balanced equation becomes crucial when calculating the amounts of substances needed or produced, such as when determining the product yield based on available reactants.
Molar Mass Calculations
Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol), and is essential for converting quantities in chemistry. Calculating molar mass involves adding up the atomic masses of all atoms in a molecule.

In our exercise, the molar mass of \( X_2 \) was calculated using:
  • Mass of \( X_2 \) \( = 0.105\) g and number of molecules \( = 8.92 \times 10^{20} \).
  • Use Avogadro鈥檚 number \( (6.022 \times 10^{23}) \) to find moles: \( 0.105 \text{ g} / (8.92 \times 10^{20} / 6.022 \times 10^{23}) \).
  • This gave a molar mass of \( 70.81\) g/mol for \( X_2 \).
The atomic mass of \( X \) was half of \( X_2 \), since \( X_2 \) is diatomic. Therefore, \( 70.81\text{ g/mol} / 2 = 35.40 \text{ g/mol}\). Understanding these calculations helps in identifying unknown elements in compounds.
Limiting Reactants
In chemical reactions, the limiting reactant is the substance that is completely consumed first, limiting the amount of products formed. Identifying the limiting reactant is key to predicting reaction yields.

In the exercise:
  • Two reactants were available: 1.00 g each of \( M \) and \( X_2 \).
  • You compute moles for each: \( \text{moles of } M = \frac{1.00 \text{ g}}{88.75 \text{ g/mol}} = 0.01127 \text{ moles} \) and \( \text{moles of } X_2 = \frac{1.00 \text{ g}}{70.81 \text{ g/mol}} = 0.01412 \text{ moles} \).
  • The reaction ratio is 1:1, making \( M \) the limiting reactant because it produces less product considering mole-to-mole comparison.
The moles of limiting reactant determine the amount of \( MX_3 \) that can be produced. Calculating this involves multiplying the moles of limiting reactant by the formula weight of the desired product, resulting in the mass of the product formed.

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

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH (188.35 g/mol) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{g} / \mathrm{mol})\) d. \(\mathrm{SN}(184.32 \mathrm{g} / \mathrm{mol})\)

In the production of printed circuit boards for the electronics industry, a 0.60-mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96 \mathrm{g} / \mathrm{cm}^{3}\)). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Natural rubidium has the average mass of 85.4678 u and is composed of isotopes \(^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \(^{87} \mathrm{Rb}\). The ratio of atoms \(^{85} \mathrm{Rb} /^{87} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has an LD\(_{50}\) (the amount of substance that is lethal to \(50 . \%\) of a population sample) of \(10 . \mu g\) per \(\mathrm{kg}\) of body mass. Tetrodotoxin is \(41.38 \%\) carbon by mass, \(13.16\%\) nitrogen by mass, and \(5.37\%\) hydrogen by mass, with the remaining amount consisting of oxygen. What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21} \mathrm{g},\) what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the \(\mathrm{LD}_{50}\) dosage for a person weighing 165 lb?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

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.