Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) \(6 \mathrm{HNO}_{3}+\mathrm{S} \longrightarrow 6 \mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) (b) \(2 \mathrm{CrCl}_{b}+3 \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{CrO}_{4}\) (c) \(\begin{aligned} 2 \mathrm{KMnO}_{4}+6 \mathrm{HCl}+& 5 \mathrm{H}_{2} \mathrm{~S} \rightarrow \\ & 2 \mathrm{KCl}+2 \mathrm{MnCl}_{2}+5 \mathrm{~S}+8 \mathrm{H}_{2} \mathrm{O} \end{aligned}\)

Short Answer

Expert verified
(a) Balanced. (b) After correcting a potential typo (CrClb to CrCl3) and adjusting water and hydrogen chloride molecules, the balanced equation is \(2 \mathrm{CrCl}_{3} + 3 \mathrm{MnO}_{2} + 8 \mathrm{H}_{2}O \rightarrow 3 \mathrm{MnCl}_{2} + 2 \mathrm{Cr}(\mathrm{OH})_{3} + 8 \mathrm{HCl}\). (c) Balanced.

Step by step solution

01

- Balancing Equation (a)

Whenever you are asked to balance a chemical equation, you should ensure that the number of atoms of each type is the same on both sides of the equation. For equation (a), no balancing is necessary as the equation is already balanced: on both sides there are 6 nitrogen atoms, 1 sulfur atom, 8 hydrogen atoms, and 18 oxygen atoms.
02

- Identifying the Error in Equation (b)

Inspect equation (b) carefully. Notice that the subscript 'b' next to CrCl seems to be a typo since typically this would be an integer representing the number of chloride ions. Confirm what the correct chemical formula should be, most likely it is a typo and should be CrCl3 instead of CrClb.
03

- Balancing Equation (b) after Correction

Assuming that the corrected formula is CrCl3, balance the revised equation: \(2 \mathrm{CrCl}_{3} + 3 \mathrm{MnO}_{2} + 8 \mathrm{H}_{2}O \rightarrow 3 \mathrm{MnCl}_{2} + 2 \mathrm{Cr}(\mathrm{OH})_{3} + 8 \mathrm{HCl}\). Each element type has the same count on each side: 2 chromium, 6 chlorine, 3 manganese, 14 hydrogen, and 10 oxygen atoms.
04

- Balancing Equation (c)

For equation (c), count the atoms of each type to make sure they are balanced on both sides. The equation is already balanced: 2 potassium, 2 manganese, 6 sulfur, 20 hydrogen, and 4+12=16 oxygen atoms.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction
When substances interact to form new products, a chemical reaction occurs. This process involves breaking chemical bonds in reactants and forming new bonds in products. It's essential to understand that during a chemical reaction, atoms are neither created nor destroyed; instead, they are rearranged to become different substances. For instance, in exercise (a), sulfur (S) and nitric acid ((HNO3)) react to form nitrogen dioxide ((NO2)), sulfuric acid ((H2SO4)), and water ((H2O)). Here, sulfur and the nitrate from nitric acid change partners - sulfur pairs with oxygen and hydrogen to form sulfuric acid, while nitrogen pairs with oxygen to form nitrogen dioxide. Recognizing the substance interchange is key to comprehending chemical reactions.
Stoichiometry
The branch of chemistry called stoichiometry deals with the quantitative relationships of substances in a chemical reaction. It's like a recipe that tells you the exact amount of each ingredient you need. Stoichiometry is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Hence, you need the same number of each type of atom on both sides of a chemical equation. In exercise (a), for example, the stoichiometry indicates that one sulfur atom reacts with six molecules of nitric acid, yielding one sulfuric acid molecule, six molecules of nitrogen dioxide, and two molecules of water. Getting the stoichiometry right means ensuring the mass of reactants equals the mass of products.
Chemical Formula Correction
Sometimes, equations may contain typos or errors that need correction before balancing, which is termed chemical formula correction. An incorrect formula can lead to an incorrect stoichiometry, making it impossible to balance the equation properly. In step 2 of equation (b), the apparent typo 'CrClb' should be 'CrCl3', which represents chromium(III) chloride. Correcting chemical formulas is crucial because each element's subscript in the formula denotes the number of atoms in a molecule, influencing the outcome of a reaction. Accurate chemical formulas ensure that the stoichiometry reflects the true nature of the chemical reaction.
Reaction Balancing Steps
Reaction balancing steps are systematic procedures used to ensure that a chemical equation adheres to the law of conservation of mass. The balancing process involves adjusting the coefficients (whole numbers placed before reactants and products) to get the same number of atoms for each element on both sides of the equation. For instance, consider equation (b) after correcting the formula; to balance it, coefficients are adjusted so that the atoms of chromium, chlorine, manganese, hydrogen, and oxygen are equal on both sides. Practicing these balancing steps improves one's ability to visualize and solve chemical equation problems, which is a foundational skill for any chemistry student.

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

There is socsething incorrect about these half-reactions: (a) \(\mathrm{Cu}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Cu}^{2+}\) (b) \(\mathrm{Pb}^{2+}+\mathrm{e}^{2-} \longrightarrow \mathrm{Pb}\) Identify what is wrong and correct at.

Differentiate between an electrolytic cell and a voltaic cell.

The following observations were made concerning metals \(A, B, C\), and \(D\). (a) When a strip of metal \(\mathrm{A}\) is placed in a solution of \(\mathrm{B}^{2+}\) aons, no reaction is observed. (b) Similarly, \(A\) in a solution containing \(\mathrm{C}^{+}\)ions produces no reaction. (c) When a strip of metal \(D\) is placed in a solut?on of " \(\mathrm{C}^{+}\)àns, black metallic C deposits on the surface of D, and the solution tests posatively for \(\mathrm{D}^{2+}\) tons. (d) When a piece of metallic B is placed in a solution of \(D^{2+}\) ions, metallic \(D\) appears on the surface of \(\mathrm{B}\) and \(\mathrm{B}^{2+}\) jons are found in the solution. Arrange the ions \(\mathrm{A}^{4}, \mathrm{~B}^{2+}, \mathrm{C}^{4}\), and \(\mathrm{D}^{24}\) in order of their ability to attract clectrons. List them in order of increasing ability.

Why is a porous barrier or a salt bridge necessary in some voltaxe cells?

Use this unbalanced redox equation $$ \mathrm{KMnO}_{4}+\mathrm{HCl} \longrightarrow \mathrm{KCl}+\mathrm{MnCl}_{2}+\mathrm{H}_{2}+\mathrm{Cl}_{2} $$ to indicate (a) the oxidizing agent (b) the reducing agent (c) the number of electrons that are transferred per mole of oxid izing agcnt

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