Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078131615
Author: Martin Silberberg Dr.
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3.4, Problem 3.20BFP
Interpretation Introduction

Interpretation:

The mass of carbon dioxide produced when 4.65 g of butane is burned in 10.0 g of oxygen is to be calculated. Also, the mass of excess reactant that remains unreacted when the reaction is over is to be calculated.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount(mol) of excess reactant is reactant left after the formation of maximum amount(mol) of products.

The formula to calculate moles is as follows:

  Amount(mol)=massmolar mass        (1)

Expert Solution & Answer
Check Mark

Answer to Problem 3.20BFP

The mass of carbon dioxide produced is 8.46 g and the mass of excess reactant that remains unreacted is 1.86g.

Explanation of Solution

The balanced equation for the combustion of butane is as follows:

  2C4H10(g)+13O2(s)8CO2(s)+10H2O

Substitute 4.65 g for mass and 58.12 g/mol for molar mass in equation (1) to calculate amount (mol) of C4H10.

  Moles of C4H10=4.65 g58.12 g/mol=0.080006 mol

Substitute 10 g for mass and 32.00 g/mol for molar mass mass in equation (1) to calculate amount (mol) of O2.

  Moles of O2=10 g32.00 g/mol=0.3125 mol

According to the balanced chemical equation, the stoichiometric ratio between C4H10 and CO2 is 2:8. Therefore, the number of moles of CO2 is calculated as:

  Amount of CO2=(0.080006 mol)(8 mol CO22 mol C4H10)=0.640055 mol

According to the balanced chemical equation, the stoichiometric ratio between O2 and CO2 is 13:8. Therefore, the number of moles of CO2 is calculated as:

  Amount of CO2=(0.3125 mol)(8 mol O213 mol CO2)=0.192307 mol

O2 is the limiting reactant, as it produces less amount of product CO2.

The formula to calculate the mass is as follows:

  Mass=(amount(mol))(molecular mass)        (2)

Substitute 0.192307 mol for amount(mol) and 44.01 g/mol for molecular mass in equation (2) to calculate mass of CO2.

  Mass of CO2=(0.192307 mol)(44.01 g/mol)=8.4635 g8.46 g

The remaining mass of the excess reagent can be calculated from the amount of CO2 combining with the limiting reagent.

  Moles of C4H10 required=[(0.3125 mol)(2 molC4H1013 molAl)]=0.0480769 mol

Substitute 0.0480769 mol for amount (mol) and 58.12 g/mol for molecular mass in equation (2) to calculate the excess amount (mol) of C4H10.

  Mass of C4H10=(0.0480769 mol)(58.12 g/mol)=2.7942 g2.79g

The excess mass of C4H10 is calculated as follows:

  Excess mass of C4H10=Initial massreacted mass        (3)

Substitute 4.65 g for initial mass and 2.79g for reacted mass in the equation (3).

  Excess mass of Al=4.65 g2.79g=1.86g.

Conclusion

The limiting reagent controls the amount of the other reactants and the product.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 3 Solutions

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.75PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.77PCh. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Prob. 3.95PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.97PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Prob. 3.101PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.104PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.107PCh. 3 - Prob. 3.108PCh. 3 - Prob. 3.109PCh. 3 - Prob. 3.110PCh. 3 - Prob. 3.111PCh. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - Prob. 3.121PCh. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.128PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Prob. 3.137PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.140P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY