/*! 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 62 The reaction between potassium s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 62

The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in self-contained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced?

Short Answer

Expert verified
(a) \(0.300\, \mathrm{mol}\, \mathrm{O}_{2}\) are produced when \(0.400\, \mathrm{mol}\, \mathrm{KO}_{2}\) reacts. (b) \(22.2 \mathrm{g}\) of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{g}\) of \(\mathrm{O}_{2}\). (c) \(6.86 \mathrm{g}\) of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{g}\) of \(\mathrm{O}_{2}\) are produced.

Step by step solution

01

(a): Find the moles of O鈧� produced

We are given that \(0.400\, \mathrm{mol}\) of \(\mathrm{KO}_{2}\) reacts. From the balanced chemical equation, we have: $$\frac{3\, \mathrm{mol}\, \mathrm{O}_{2}}{4\, \mathrm{mol}\, \mathrm{KO}_{2}}$$ So, to find the moles of oxygen produced, we multiply the moles of \(\mathrm{KO}_{2}\) by the stoichiometric ratio: $$0.400\, \mathrm{mol}\, \mathrm{KO}_{2} \times \frac{3\, \mathrm{mol}\, \mathrm{O}_{2}}{4\, \mathrm{mol}\, \mathrm{KO}_{2}}$$ Let's calculate the moles of oxygen produced: $$0.400\, \mathrm{mol}\, \mathrm{KO}_{2} \times \frac{3}{4} = 0.300\, \mathrm{mol}\, \mathrm{O}_{2}$$ So, there are \(0.300\, \mathrm{mol}\) of \(\mathrm{O}_{2}\) produced.
02

(b): Find the grams of KO鈧� needed

We have to find the grams of \(\mathrm{KO}_{2}\) needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). First, we need to find the moles of \(\mathrm{O}_{2}\) in \(7.50 \mathrm{~g}\). To do this, we will use the molar mass of oxygen: $$\frac{1\,\mathrm{mol}\, \mathrm{O}_{2}}{32.00\, \mathrm{g}\, \mathrm{O}_{2}}$$ Calculate the moles of oxygen: $$\frac{7.50\, \mathrm{g}\, \mathrm{O}_{2}}{32.00\, \mathrm{g/mol}} = 0.234\, \mathrm{mol}\, \mathrm{O}_{2}$$ Now, we will use the stoichiometric ratio from the balanced chemical equation: $$\frac{4\, \mathrm{mol}\, \mathrm{KO}_{2}}{3\, \mathrm{mol}\, \mathrm{O}_{2}}$$ Calculate the moles of \(\mathrm{KO}_{2}\) needed: $$0.234 \, \mathrm{mol}\, \mathrm{O}_{2} \times \frac{4}{3} = 0.312\, \mathrm{mol}\, \mathrm{KO}_{2}$$ Finally, we use the molar mass of \(\mathrm{KO}_{2}\) to find the grams of \(\mathrm{KO}_{2}\): $$\frac{1\, \mathrm{mol}\, \mathrm{KO}_{2}}{71.10\, \mathrm{g}\, \mathrm{KO}_{2}}$$ Calculate the grams of \(\mathrm{KO}_{2}\) needed: $$0.312\, \mathrm{mol}\, \mathrm{KO}_{2} \times 71.10\, \mathrm{g/mol} = 22.2\, \mathrm{g}\, \mathrm{KO}_{2}$$ So, \(22.2 \mathrm{~g}\) of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\).
03

(c): Find the grams of CO鈧� used

We know from (b) that \(0.234\, \mathrm{mol}\) of \(\mathrm{O}_{2}\) are produced from \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). We will use the stoichiometric ratio from the balanced chemical equation to find the moles of \(\mathrm{CO}_{2}\): $$\frac{2\, \mathrm{mol}\, \mathrm{CO}_{2}}{3\, \mathrm{mol}\, \mathrm{O}_{2}}$$ Calculate the moles of \(\mathrm{CO}_{2}\) used: $$0.234\, \mathrm{mol}\, \mathrm{O}_{2} \times \frac{2}{3} = 0.156\, \mathrm{mol}\, \mathrm{CO}_{2}$$ Finally, we use the molar mass of \(\mathrm{CO}_{2}\) to find the grams of \(\mathrm{CO}_{2}\): $$\frac{1\, \mathrm{mol}\, \mathrm{CO}_{2}}{44.01\, \mathrm{g}\, \mathrm{CO}_{2}}$$ Calculate the grams of \(\mathrm{CO}_{2}\) used: $$0.156\, \mathrm{mol}\, \mathrm{CO}_{2} \times 44.01\, \mathrm{g/mol} = 6.86\, \mathrm{g}\, \mathrm{CO}_{2}$$ So, \(6.86 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced.

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.

Understanding Chemical Equations
A chemical equation is a symbolic representation of a chemical reaction. It describes the substances involved and their quantities. In the provided reaction between potassium superoxide \( \text{KO}_2 \) and carbon dioxide \( \text{CO}_2 \), the equation is shown as:\[4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} \] This tells us several important things:
  • Each side of the equation must be balanced in terms of the number of each type of atom. This means that there are equal numbers of potassium (K), oxygen (O), and carbon (C) atoms on both sides.
  • The coefficients (numbers in front of molecules like 4, 2, and 3) indicate the ratio between the reactants and products.
  • Only the coefficients can be changed to balance the equation, not the subscripts within the chemical formulas.
Chemical equations help us understand the specific relationships and proportions of chemicals that are consumed and produced in a chemical reaction.
Exploring Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (\text{g/mol}). It's calculated by summing the atomic masses of all the atoms in a compound's formula.For example, the molar mass of oxygen gas \( \text{O}_2 \) is calculated from the atomic mass of an oxygen atom (approximately 16 \, \text{g/mol}). Since there are two oxygen atoms in \( \text{O}_2,\) its molar mass is: \[ 2 \times 16.00 = 32.00 \, \text{g/mol} \]Similarly, the molar mass of \( \text{KO}_2 \) is calculated by adding the atomic masses of potassium (\approx 39.10)\ and two oxygen atoms (2 \times 16.00):\[ 39.10 + 2 \times 16.00 = 71.10 \, \text{g/mol} \]Knowing the molar mass helps in converting between grams and moles, which is essential when dealing with chemical reactions and stoichiometry.
Reactant-Product Relationships in Reactions
In any chemical reaction, understanding the relationship between reactants and products is key to solving stoichiometry problems. Stoichiometry itself is the quantitative study of these relationships.From the balanced equation, \( 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2},\) we can deduce:
  • 4 moles of \( \mathrm{KO}_2\) react with 2 moles of \( \mathrm{CO}_2 \) to produce 3 moles of \( \mathrm{O}_2 \).
  • These coefficients provide the ratios used in stoichiometric conversions. For instance, \(\frac{3 \, \mathrm{mol} \, \mathrm{O}_2}{4 \, \mathrm{mol} \, \mathrm{KO}_2}\) helps in determining how many moles of \( \mathrm{O}_2 \) are produced from a given number of moles of \( \mathrm{KO}_2 \).
  • Using stoichiometric ratios, one can calculate the required amount of reactants or the expected quantity of products.
Understanding these relationships ensures the efficient calculation and effective prediction of chemical reaction outcomes.

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

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from \(14.2 \mathrm{~g}\) of aluminum sulfide?

(a) You are given a cube of silver metal that measures \(1.000 \mathrm{~cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(10.4 \% \mathrm{C}, 27.8 \% \mathrm{~S},\) and \(61.7 \% \mathrm{Cl}\) (b) \(21.7 \% \mathrm{C}, 9.6 \% \mathrm{O},\) and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al},\) and the remainder \(\mathrm{F}\)

(a) When the metallic element sodium combines with the nonmetallic element bromine, \(\mathrm{Br}_{2}(l),\) how can you determine the chemical formula of the product? How do you know whether the product is a solid, liquid, or gas at room temperature? Write the balanced chemical equation for the reaction. (b) When a hydrocarbon burns in air, what reactant besides the hydrocarbon is involved in the reaction? What products are formed? Write a balanced chemical equation for the combustion of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) in air.

If Avogadro's number of pennies is divided equally among the 300 million men, women, and children in the United States, how many dollars would each receive? How does this compare with the gross domestic product (GDP) of the United States, which was \(\$ 14.4\) trillion in \(2008 ?\) (The GDP is the total market value of the nation's goods and services.)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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.