Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 115

A mixture contains only \(\mathrm{CuCl}_{2}\) and \(\mathrm{FeCl}_{3}\). A \(0.7391 \mathrm{g}\) sample of the mixture is completely dissolved in water and then treated with \(\mathrm{AgNO}_{3}(\) aq). The following reactions occur. $$\begin{aligned} &\mathrm{CuCl}_{2}(\mathrm{aq})+2 \mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow 2 \mathrm{AgCl}(\mathrm{s})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \end{aligned}$$ $$\begin{aligned} &\mathrm{FeCl}_{3}(\mathrm{aq})+3 \mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow 3 \mathrm{AgCl}(\mathrm{s})+\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(\mathrm{aq})\end{aligned}$$ If it takes \(86.91 \mathrm{mL}\) of \(0.1463 \mathrm{M} \mathrm{AgNO}_{3}\) solution to precipitate all the chloride as \(\mathrm{AgCl}\), then what is the percentage by mass of copper in the mixture?

Short Answer

Expert verified
As the percentage by mass of copper will depend on the calculated values from the steps above, the exact value cannot be given here. However, the short answer would be the final percentage value obtained from step 4.

Step by step solution

01

Calculate moles of \(\mathrm{AgNO}_{3}\) used

Firstly, calculate the moles of \(\mathrm{AgNO}_{3}\) used using the formula \(n = V \times c\), where \(n\) is the number of moles, \(V\) is the volume in liters and \(c\) is the concentration in molarity (mol/L). Substituting the given volume and concentration, the calculation becomes: \(n = 0.08691 \times 0.1463\)
02

Determine moles of chloride ions

Next, determine the total moles of chloride ions. As both reactions have AgCl as a product (and each AgCl is formed from one Cl- ion), the total moles of chloride ions is equal to the total moles of \(\mathrm{AgNO}_{3}\) used.
03

Determine moles and mass of copper

Knowing that each mole of \(\mathrm{CuCl}_{2}\) contributes 2 moles of chloride ions, the moles of \(\mathrm{CuCl}_{2}\) will be half of the total moles of chloride ions. The molar mass of copper (Cu) is 63.546 g/mol. To determine the mass of copper, multiply the moles of \(\mathrm{CuCl}_{2}\) by the molar mass of copper.
04

Calculate percentage by mass of copper

Finally, calculate the percentage by mass of copper in the mixture by dividing the mass of copper by the total mass of the mixture (in this case, 0.7391 g) and multiply by 100.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Which has the higher concentration of sucrose: a \(46 \%\) sucrose solution by mass \((d=1.21 \mathrm{g} / \mathrm{mL}),\) or \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ? Explain your reasoning.

It is often difficult to determine the concentration of a species in solution, particularly if it is a biological species that takes part in complex reaction pathways. One way to do this is through a dilution experiment with labeled molecules. Instead of molecules, however, we will use fish. An angler wants to know the number of fish in a particular pond, and so puts an indelible mark on 100 fish and adds them to the pond's existing population. After waiting for the fish to spread throughout the pond, the angler starts fishing, eventually catching 18 fish. Of these, five are marked. What is the total number of fish in the pond?

A 0.696 mol sample of \(\mathrm{Cu}\) is added to \(136 \mathrm{mL}\) of \(6.0 \mathrm{M}\) HNO \(_{3}\). Assuming the following reaction is the only one that occurs, will the Cu react completely? $$\begin{aligned} 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) +4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{NO}(\mathrm{g}) \end{aligned}$$

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Chromium(II) sulfate, \(\mathrm{CrSO}_{4^{\prime}}\) is a reagent that has been used in certain applications to help reduce carbon-carbon double bonds \((\mathrm{C}=\mathrm{C})\) in molecules to single bonds ( \(\mathrm{C}-\mathrm{C}\) ). The reagent can be prepared via the following reaction. $$\begin{array}{c} 4 \mathrm{Zn}(\mathrm{s})+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{ZnSO}_{4}(\mathrm{aq})+2 \mathrm{CrSO}_{4}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

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.

Sign-up for free