Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 65

What mass of solid AgBr is produced when 100.0 \(\mathrm{mL}\) of 0.150 \(\mathrm{MAgNO}_{3}\) is added to 20.0 \(\mathrm{mL}\) of 1.00 $\mathrm{M} \mathrm{NaBr} ?$

Short Answer

Expert verified
The mass of solid AgBr produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0 mL of 1.00 M NaBr is approximately 2.82 g.

Step by step solution

01

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between AgNO3 and NaBr to obtain AgBr and NaNO3. The balanced chemical equation is as follows: \[AgNO_{3}(aq) + NaBr(aq) \rightarrow AgBr(s) + NaNO_{3}(aq)\]
02

Find moles of each reactant

To find the limiting reactant, we must first determine the moles of each reactant present in the reaction. We will use the given volumes and concentrations to calculate the number of moles for each reactant. For AgNO3: Moles = (volume in L) × (concentration in M) Moles of AgNO3 = (100.0 mL × (1 L/1000 mL)) × (0.150 mol/L) Moles of AgNO3 = 0.015 mol For NaBr: Moles = (volume in L) × (concentration in M) Moles of NaBr = (20.0 mL × (1 L/1000 mL)) × (1.00 mol/L) Moles of NaBr = 0.020 mol
03

Identify the limiting reactant

Now, we will compare the moles of reactants to determine the limiting reactant. The stoichiometry of the balanced chemical equation shows us that 1 mole of AgNO3 reacts with 1 mole of NaBr. From the calculated moles of reactants: 0.015 mol AgNO3 × (1 mol NaBr / 1 mol AgNO3) = 0.015 mol NaBr Since we have 0.020 mol of NaBr, but only 0.015 mol are required to react with the available AgNO3, AgNO3 is the limiting reactant. The reaction will stop when all 0.015 mol of AgNO3 have been consumed.
04

Calculate the mass of AgBr produced

Since AgNO3 is the limiting reactant, the amount of AgBr produced will depend on the moles of AgNO3. We will use the stoichiometry of the balanced chemical equation to find the moles of AgBr produced. Moles of AgBr produced = (moles of limiting reactant AgNO3) × (1 mol AgBr / 1 mol AgNO3) Moles of AgBr produced = 0.015 mol Now, we will convert moles of AgBr to mass. To do this, we need to know the molar mass of AgBr, which is approximately 187.77 g/mol. Mass of AgBr = (moles of AgBr) × (molar mass of AgBr) Mass of AgBr = (0.015 mol) × (187.77 g/mol) Mass of AgBr = 2.82 g The mass of solid AgBr produced is approximately 2.82 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.

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

Many oxidation–reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. $\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)$ b. $\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)$ c. $\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(i)$ d. $\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$

Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and 4.09% H by mass. A titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 g carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?

What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH? a. 0.100 \({M} {HCl} \quad \)c. 0.200$M {HC}_{2} {H}_{3} {O}_{2}$$(1 \text { acidic hydrogen })$ b. 0.150 \({M} {HNO}_{3}\)

A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The BaSO\(_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Organic 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.

Sign-up for free