Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 64

You are given a \(1.50-\mathrm{g}\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{~mL}\) of water and then add an excess of \(0.500 \mathrm{M}\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of \(0.641 \mathrm{~g}\). a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.

Short Answer

Expert verified
The species in the solution are Na⁺, NO₃⁻, Cl⁻, and Ag⁺. The balanced net ionic equation for the reaction producing the solid is Cl⁻ (aq) + Ag⁺ (aq) → AgCl (s). The percent of sodium chloride in the original unknown mixture is approximately 17.4%.

Step by step solution

01

Identify dissolved species in water

After dissolving the mixture of sodium nitrate (NaNO₃) and sodium chloride (NaCl) in water, the species present are sodium ions (Na⁺), nitrate ions (NO₃⁻), and chloride ions (Cl⁻). When the silver nitrate (AgNO₃) solution is added, we also have silver ions (Ag⁺) in the solution. So, the species we would see in the solution are Na⁺, NO₃⁻, Cl⁻, and Ag⁺. b. Write the balanced net ionic equation
02

Write complete ionic equation

In the complete ionic equation, all soluble ionic compounds are shown as individual ions. In our case, the equation is: Na⁺ (aq) + NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) + Ag⁺ (aq) + NO₃⁻ (aq) → Na⁺ (aq) + NO₃⁻ (aq) + AgCl (s) + Na⁺ (aq) + NO₃⁻ (aq)
03

Cross out the spectator ions

Spectator ions are the ions that do not participate in the reaction. In this case, Na⁺ and NO₃⁻ are the spectator ions. Crossing them out, we get: Cl⁻ (aq) + Ag⁺ (aq) → AgCl (s) b. Balanced net ionic equation:
04

Balanced net ionic equation for the reaction

The balanced net ionic equation for the reaction that produces the solid is: Cl⁻ (aq) + Ag⁺ (aq) → AgCl (s) c. Calculate the percent sodium chloride in the original mixture
05

Find the mass of silver chloride formed

The mass of the white solid (silver chloride) formed is 0.641 g.
06

Use stoichiometry to find the mass of sodium chloride

Using the stoichiometry of the balanced net ionic equation, we can find the mass of sodium chloride that reacted to form silver chloride. The molar mass of AgCl = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol The molar mass of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol 1 mol of NaCl reacts with 1 mol of AgCl. So, mass of NaCl = (mass of AgCl / molar mass of AgCl) * molar mass of NaCl = (0.641 g / 143.32 g/mol) * 58.44 g/mol ≈ 0.261 g
07

Calculate the percent sodium chloride in the original mixture

The total mass of the original mixture is given as 1.50 g. The mass of sodium chloride we found is 0.261 g. Percent NaCl = (mass of NaCl / total mass of the mixture) * 100 Percent NaCl = (0.261 g / 1.50 g) * 100 ≈ 17.4 % c. The percent sodium chloride in the original unknown mixture is approximately 17.4 %.

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

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584 \mathrm{~g}\) pure manganese metal in nitric acid and diluting to a final volume of \(1.000 \mathrm{~L}\). The following solutions were then prepared by dilution: For solution A. \(50.00 \mathrm{~mL}\) of stock solution was diluted to \(1000.0 \mathrm{~mL}\) For solution \(B, 10.00 \mathrm{~mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{~mL}\). For solution \(C, 10.00 \mathrm{~mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{~mL}\). Calculate the concentrations of the stock solution and solutions \(A, B\), and \(C\).

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the \(29.0-\mathrm{g}\) mixture.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

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