/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Write balanced equations for the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 65

Write balanced equations for the following reactions: (a) barium oxide with water, (b) iron(II) oxide with perchloric acid, (c) sulfur trioxide with water, (d) carbon dioxide with aqueous sodium hydroxide.

Short Answer

Expert verified
(a) \(BaO + H_2O \rightarrow Ba(OH)_2\) (b) \(FeO + 2HClO_4 \rightarrow Fe(ClO_4)_2 + H_2O\) (c) \(SO_3 + H_2O \rightarrow H_2SO_4\) (d) \(CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O\)

Step by step solution

01

Understanding the reactants and predicting the products

We need to first identify the reactants given in each reaction and predict the products formed. Each reaction is described below: (a) Barium oxide (BaO) reacts with water (H2O) to form barium hydroxide (Ba(OH)2). (b) Iron(II) oxide (FeO) reacts with perchloric acid (HClO4) to form iron(II) perchlorate (Fe(ClO4)2) and water (H2O). (c) Sulfur trioxide (SO3) reacts with water (H2O) to form sulfuric acid (H2SO4). (d) Carbon dioxide (CO2) reacts with aqueous sodium hydroxide (NaOH) to form sodium carbonate (Na2CO3) and water (H2O). Next, we need to balance these equations to satisfy the law of conservation of mass.
02

Balancing the equations

(a) BaO + H2O -> Ba(OH)2 The equation is already balanced, with one Ba, two O, and two H atoms on each side. (b) FeO + 2HClO4 -> Fe(ClO4)2 + H2O We need to balance the Cl and O atoms. There are four O atoms and two Cl atoms on the right side, so we put a coefficient of 2 in front of HClO4. Now the equation is balanced with one Fe, two Cl, and six O atoms on each side. (c) SO3 + H2O -> H2SO4 This equation is already balanced, with two H, one S, and four O atoms on each side. (d) CO2 + 2NaOH -> Na2CO3 + H2O We need to balance the Na atoms on both sides of the equation. We put a coefficient of 2 in front of NaOH, making the equation balanced, with one C, three O, and two Na atoms on each side. Here are the balanced equations for the four given reactions: (a) BaO + H2O -> Ba(OH)2 (b) FeO + 2HClO4 -> Fe(ClO4)2 + H2O (c) SO3 + H2O -> H2SO4 (d) CO2 + 2NaOH -> Na2CO3 + H2O

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 91影视!

Key Concepts

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

Chemical Reaction Prediction
Predicting the products of chemical reactions is an essential skill in chemistry. It involves understanding how different substances interact based on their chemical properties and the types of bonds they form. For instance, when a metal oxide reacts with water, it generally forms a metal hydroxide, as shown in the reaction between barium oxide and water; BaO + H2O -> Ba(OH)2. In the case of a reaction between an oxide and an acid, like iron(II) oxide with perchloric acid, the products are typically a salt and water, leading to the equation FeO + 2HClO4 -> Fe(ClO4)2 + H2O.

When dealing with more complex compounds like sulfur trioxide and water, it's known that they form an acid, specifically sulfuric acid in this case: SO3 + H2O -> H2SO4. A keen understanding of the reactants' chemical nature helps us to predict that carbon dioxide and sodium hydroxide will yield sodium carbonate and water: CO2 + 2NaOH -> Na2CO3 + H2O. Through practice and studying different types of reactions, students can improve their ability to predict reaction products more accurately.
Stoichiometry
Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is founded upon the balanced chemical equation, which provides the ratio in which substances react and form products. To apply stoichiometry to balance equations, you look at the moles of each element and ensure the same number of atoms of each element appears on both sides of the equation. For instance, in balancing the reaction CO2 + 2NaOH -> Na2CO3 + H2O, we notice that there are two sodium atoms in the sodium carbonate and therefore require two NaOH molecules to balance the equation.

Stoichiometry also allows you to calculate the amount of product formed from given quantities of reactants or vice versa, making it deeply related to practical applications such as determining the yield of a reaction or calculating the necessary amounts of reactants for a complete reaction.
Law of Conservation of Mass
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. It is the fundamental principle guiding the balancing of chemical equations. During a chemical reaction, the atoms rearrange, but the total mass of the system remains constant. This is why in balancing the equations provided, attention was paid to ensure that the number of each type of atom on the reactant side is equivalent to that on the product side.

For example, when balancing the equation FeO + 2HClO4 -> Fe(ClO4)2 + H2O, it's crucial to ensure that the atoms of iron, chlorine, and oxygen are conserved. Here, there's one iron, two chlorines, and a total of six oxygens on both sides after balancing. Emphasizing the concept of conservation of mass can significantly aid in understanding why chemical equations must be balanced and how this conservation underpins the quantitative aspect of chemical reactions.

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

Detailed calculations show that the value of \(Z_{\text { eff }}\) for the outermost electrons in Si and Cl atoms is \(4.29+\) and \(6.12+\) , respectively.(a) What value do you estimate for \(Z\) eff experienced by the outermost electron in both Si and Cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text { eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text { eff? }} ?(\mathbf{d})\) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text { eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text { eff }}\) for a valence electron in \(\mathrm{P}\) , phosphorus, based on the calculations for Si and Cl.

Give three examples of ions that have an electron configuration of \(n d^{6}(n=3,4,5, \ldots).\)

Consider the isoelectronic ions \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+} .\) (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, \(S,\) calculate \(Z_{\text { eff }}\) for the 2\(p\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S\) .(d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

We will see in Chapter 12 that semiconductors are materials that conduct electricity better than nonmetals but not as well as metals. The only two elements in the periodic table that are technologically useful semiconductors are silicon and germanium. Integrated circuits in computer chips today are based on silicon. Compound semiconductors are also used in the electronics industry. Examples are gallium arsenide, GaAs; gallium phosphide, GaP; cadmium sulfide, CdS; and cadmium selenide, CdSe. (a) What is the relationship between the compound semiconductors鈥� compositions and the positions of their elements on the periodic table relative to Si and Ge? (b) Workers in the semiconductor industry refer to "II鈥揤I" and "III鈥揤" materials, using Roman numerals. Can you identify which compound semiconductors are II鈥揤I and which are III鈥揤? (c) Suggest other compositions of compound semiconductors based on the positions of their elements in the periodic table.

Potassium superoxide, \(\mathrm{KO}_{2},\) is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that 2 \(\mathrm{mol}\) of \(\mathrm{KO}_{2}(s)\) react with each mole of= \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume 18.0 \(\mathrm{g} \mathrm{CO}_{2}(g) ?\) What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

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.