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. $$\begin{array}{c} 4 \mathrm{Zn}(\mathrm{s})+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{ZnSO}_{4}(\mathrm{aq})+2 \mathrm{CrSO}_{4}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$

Short Answer

Expert verified
The chemical reaction involves the reactants zinc (Zn), \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \) that react to produce \( \mathrm{ZnSO}_{4} \), \( \mathrm{CrSO}_{4} \), \( \mathrm{K}_{2} \mathrm{SO}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \). The equation is balanced, having equal numbers of each atom on both sides.

Step by step solution

01

Identify the reactants and products

The reagents for this reaction are zinc (Zn), \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \). The products of this reaction are \( \mathrm{ZnSO}_{4} \), \( \mathrm{CrSO}_{4} \), \( \mathrm{K}_{2} \mathrm{SO}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \)
02

Balance the equation

The key to balancing the equation is to ensure that the number of atoms for each element is the same on both sides of the equation. In the given equation, we can see that this is already balanced correctly. For example, looking at the zinc atom, we have 4 zinc atoms on the left side and 4 zinc atoms that are part of \( \mathrm{ZnSO}_{4} \) on the right side.
03

Interpret the equation

This equation represents the chemical reaction where zinc, \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \) react to produce \( \mathrm{ZnSO}_{4} \), \( \mathrm{CrSO}_{4} \), \( \mathrm{K}_{2} \mathrm{SO}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \). \( \mathrm{CrSO}_{4} \) is the reagent in focus and it is used to reduce double carbon bonds to single bonds in molecules.

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 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

A small piece of zinc is dissolved in \(50.00 \mathrm{mL}\) of \(1.035 \mathrm{M}\) HCl. At the conclusion of the reaction, the concentration of the \(50.00 \mathrm{mL}\) sample is redetermined and found to be \(0.812 \mathrm{M} \mathrm{HCl} .\) What must have been the mass of the piece of zinc that dissolved? $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)

Hydrogen gas, \(\mathrm{H}_{2}(\mathrm{g}),\) is passed over \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) at \(400^{\circ} \mathrm{C} .\) Water vapor is formed together with a black residue-a compound consisting of \(72.3 \% \mathrm{Fe}\) and \(27.7 \%\) O. Write a balanced equation for this reaction.

In many communities, water is fluoridated to prevent tooth decay. In the United States, for example, more than half of the population served by public water systems has access to water that is fluoridated at approximately \(1 \mathrm{mg} \mathrm{F}^{-}\) per liter. (a) What is the molarity of \(\mathrm{F}^{-}\) in water if it contains \(1.2 \mathrm{mg} \mathrm{F}^{-}\) per liter? (b) How many grams of solid KF should be added to a \(1.6 \times 10^{8}\) L water reservoir to give a fluoride concentration of \(1.2 \mathrm{mg} \mathrm{F}^{-}\) per liter?

See all solutions

Recommended explanations on Chemistry Textbooks

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