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

The following are actual student responses to the question: Why is it necessary to balance chemical equations? a. The chemicals will not react until you have added the correct mole ratios. b. The correct products will not be formed unless the right amount of reactants have been added. c. A certain number of products cannot be formed without a certain number of reactants. d. The balanced equation tells you how much reactant you need and allows you to predict how much product you'll make. e. A mole-to-mole ratio must be established for the reaction to occur as written. Justify the best choice, and for choices you did not pick, explain what is wrong with them.

Short Answer

Expert verified
The best choice is (d), as it correctly explains the significance of balancing chemical equations: it allows us to know how much reactant is needed and predict the amount of product formed while adhering to the Law of Conservation of Mass. The other responses are not as precise or accurate in their explanations: a. Incorrect - Reactants can still react without correct mole ratios. b. Partially correct but not the main reason - The right amount of reactants may influence product formation but doesn't guarantee correct products. c. Vague - The statement does not clearly explain the importance of balancing equations. e. Not the main reason - The statement focuses on the mole-to-mole ratio without emphasizing the importance of conserving atoms in a reaction.

Step by step solution

01

Understanding the importance of balancing chemical equations

Balancing chemical equations is essential because it follows the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, in a balanced chemical equation, the same number of each type of atom must be found in both the reactants and the products.
02

Evaluating each response

a. Although mole ratios play a role in reactions, the statement is not true as correct mole ratios do not ensure reaction, and chemicals can still react without them. b. This statement is partially correct but not the main reason for balancing equations. The right amount of reactants can influence product formation, but it doesn't guarantee correct products. c. There is a relationship between the number of reactants and products, but the statement is too vague to explain the importance of balancing equations. d. This response correctly highlights that a balanced equation provides information on the required reactant amount and predictions of product formation. e. Establishing a mole-to-mole ratio is essential for a reaction to occur as written, but reactants can still react in different ratios, and this statement doesn't emphasize the importance of conserving atoms in a reaction.
03

Justifying the best choice and providing explanations for the other responses

The best choice is (d), as it correctly explains the significance of balancing chemical equations: it allows us to know how much reactant is needed and predict the amount of product formed while adhering to the Law of Conservation of Mass. The other responses are not as precise or accurate in their explanations: a. Incorrect - Reactants can still react without correct mole ratios. b. Partially correct but not the main reason - The right amount of reactants may influence product formation but doesn't guarantee correct products. c. Vague - The statement does not clearly explain the importance of balancing equations. e. Not the main reason - The statement focuses on the mole-to-mole ratio without emphasizing the importance of conserving atoms in a reaction.

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.

Law of Conservation of Mass
In the fascinating world of chemical reactions, balancing equations is more than a mere exercise in arithmetic—it is a fundamental expression of the Law of Conservation of Mass. This law states that matter is neither created nor destroyed in a chemical reaction.

What does this mean for students and scientists alike? Imagine you are baking cookies, and the recipe calls for exactly one cup of sugar, two cups of flour, and one egg. If you were to add more sugar or less flour, the outcome—your cookies—would not be the same as intended by the recipe. Similarly, when chemicals react, the reactants transform into products, but the total mass remains constant throughout this process. To visualize this, picture atoms as tiny building blocks. During a reaction, these blocks are rearranged to form new structures—the products—yet the number of blocks stays the same.

Why is this important?

By adhering to the Law of Conservation of Mass when balancing a chemical equation, we ensure that the atom count for each element is equal on both sides of the equation. This step is non-negotiable because it reflects the physical reality: in a closed system, the amount of matter remains constant, regardless of the changes that take place.
Chemical Reaction Stoichiometry
Stoichiometry might sound like a fancy term, but it's essentially the 'recipe' for a chemical reaction. Chemical reaction stoichiometry deals with the quantitative relationships between the reactants and products in a chemical reaction. Think of it as the science of counting and comparing the number of atoms and molecules involved.

As you engage with chemical reactions, stoichiometry acts as a GPS, guiding you to understand exactly how much of each reactant you’ll need and what amount of product you'll get. To put it plainly, if the stoichiometry is off, you might end up with leftover reactants, or worse, run out before the reaction is complete, very much like running out of flour halfway through baking a cake. In the context of the educational exercise, balancing equations using stoichiometry is akin to ensuring you have the right proportions in a recipe so that the ingredients combine perfectly and yield the desired product without any waste.

Real-World Applications

Stoichiometry isn't just about getting homework right; it's key in industries for maximizing product yield while minimizing waste and costs. By mastering reaction stoichiometry, students gain insights into the practical aspects of chemistry that impact pharmaceuticals, manufacturing, and environmental science.
Mole-to-Mole Ratio
The mole-to-mole ratio is a concept that might evoke images of tiny creatures, but in chemistry, a 'mole' is a unit that measures the amount of substance. The mole-to-mole ratio refers to the relative amounts of moles in reactants and products expressed in a balanced chemical equation.

This ratio is the heart of reaction stoichiometry and is critical for making precise predictions about the outcomes of chemical reactions. To understand why, consider that the balanced equation gives the proportion of reactants that react and the amount of products formed, represented in moles. Just as a chef uses cups and tablespoons to measure ingredients accurately, chemists use the mole as a measuring cup for atoms and molecules.

Applying Mole Ratios

When solving stoichiometric problems, students can use these ratios to convert between moles of one substance and moles of another. For instance, if the balanced equation tells us that 2 moles of hydrogen gas (\(H_2\)) react with 1 mole of oxygen gas (\(O_2\)) to produce 2 moles of water (\(H_2O\)), we have a clear mole-to-mole ratio to guide us. Understanding these ratios empowers students to not only balance equations but also to scale reactions up or down, such as when producing a desired amount of a chemical product in a lab or industrial setting.

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

The most common form of nylon (nylon-6) is \(63.68 \%\) carbon. \(12.38 \%\) nitrogen, \(9.80 \%\) hydrogen, and \(14.14 \%\) oxygen. Calculate the empirical formula for nylon-6.

From the information below, determine the mass of substance \(C\) that will be formed if \(45.0\) grams of substance \(A\) reacts with \(23.0\) grams of substance \(B\). (Assume that the reaction between \(A\) and \(B\) goes to completion.) a. Substance \(A\) is a gray solid that consists of an alkaline earth metal and carbon ( \(37.5 \%\) by mass). It reacts with substance \(B\) to produce substances \(C\) and \(D .\) Forty million trillion formula units of \(A\) have a mass of \(4.26\) milligrams. b. \(47.9\) grams of substance \(B\) contains \(5.36\) grams of hydrogen and \(42.5\) grams of oxygen. c. When \(10.0\) grams of \(C\) is burned in excess oxygen, \(33.8\) grams of carbon dioxide and \(6.92\) grams of water are produced. \(\mathrm{A}\) mass spectrum of substance \(C\) shows a parent molecular ion with a mass-to-charge ratio of 26 . d. Substance \(D\) is the hydroxide of the metal in substance \(A\).

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) In a certain experiment, \(20.00 \mathrm{~g}\) iron metal was reacted with \(11.20 \mathrm{~g}\) oxygen gas. After the experiment, the iron was totally consumed, and \(3.24 \mathrm{~g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

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