Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 141

In a \(1-\mathrm{L}\) beaker, \(203 \mathrm{~mL}\) of \(0.307 \mathrm{M}\) ammonium chromate was mixed with \(137 \mathrm{~mL}\) of \(0.269 M\) chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write the balanced chemical equation for the reaction occurring here. If the percent yield of the reaction was \(88.0 \%\), what mass of chromium(III) chromate was isolated?

Short Answer

Expert verified
The balanced chemical equation for the reaction is \((NH_4)_2CrO_4 + Cr(NO_2)_3 \rightarrow 2 NH_4NO_2 + Cr_2(CrO_4)_3\). The limiting reactant is chromium(III) nitrite, and the mass of chromium(III) chromate isolated is 16.83 g.

Step by step solution

01

1. Write the balanced chemical equation for the reaction.

First, let's write the chemical formulas for the reaction: Ammonium chromate: \((NH_4)_2CrO_4\) Chromium(III) nitrite: \(Cr(NO_2)_3\) Ammonium nitrite: \(NH_4NO_2\) Chromium(III) chromate: \(Cr_2(CrO_4)_3\) Now we write the balanced chemical equation: \((NH_4)_2CrO_4 + Cr(NO_2)_3 \rightarrow 2 NH_4NO_2 + Cr_2(CrO_4)_3\)
02

2. Calculate the moles of each reactant.

Next, we will calculate the number of moles for each reactant using the volume and molarity given: Moles of ammonium chromate = volume × molarity = 203 mL × 0.307 M = 0.203 L × 0.307 mol/L = 0.0623 mol Moles of chromium(III) nitrite = volume × molarity = 137 mL × 0.269 M = 0.137 L × 0.269 mol/L = 0.0368 mol
03

3. Determine the limiting reactant.

Now we need to determine the limiting reactant. We will compare the mole ratios between the reactants: Mole ratio ammonium chromate / chromium(III) nitrite = 0.0623 mol / 0.0368 mol = 1.69 From the balanced chemical equation, we can see that the mole ratio for ammonium chromate to chromium(III) nitrite should be 1:1. Since our calculated mole ratio is greater than 1, the limiting reactant is chromium(III) nitrite.
04

4. Calculate the theoretical yield of chromium(III) chromate.

Next, we determine the theoretical yield of chromium(III) chromate using the mole ratio between the limiting reactant and the product: Moles of chromium(III) chromate = moles of limiting reactant = 0.0368 mol Now, we will calculate the mass of chromium(III) chromate using the molecular weight: Molecular weight of chromium(III) chromate = 2(51.996) + 3(51.996 + 4(15.999)) = 519.86 g/mol Theoretical yield of chromium(III) chromate = moles × molecular weight = 0.0368 mol × 519.86 g/mol = 19.13 g
05

5. Calculate the actual yield of chromium(III) chromate.

We are given the percent yield of the reaction, which is 88.0%. To calculate the actual yield, we will multiply the theoretical yield by the percent yield: Actual yield of chromium(III) chromate = theoretical yield × percent yield = 19.13 g × 0.88 = 16.83 g Therefore, the mass of chromium(III) chromate isolated is 16.83g.

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

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\operatorname{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Mg} \mathrm{I}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\operatorname{AlBr}_{3}(a q)\)

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) d. As \(_{4}\) e. \(\mathrm{HAsO}_{2}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of \(0.5894 \mathrm{~g}\) were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.100\) mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{~mL}\) of solution b. \(2.5\) moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(1.25 \mathrm{~L}\) of solution c. \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{~mL}\) of solution d. \(1.00 \mathrm{~g} \mathrm{~K}_{2} \mathrm{PO}_{4}\) in \(250.0 \mathrm{~mL}\) of solution
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

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.

Sign-up for free