General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11, Problem 11.72P
Interpretation Introduction

Interpretation:

Mass of each product when 1kg of each reactant is used has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is g/mol.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  Number of moles=mass of the compoundmolar mass of the compound

Mass percent is calculated by the mass of the element divided by the mass of compound and multiplied by 100.

The formula to calculate mass percentage of substance is as follows:

  Mass % of substance=(Mass of substanceMass of compound)(100)

For more than one quantity of reactants in a reaction, the reactant that is completely consumed and controls the amount of product synthesized is called limiting reactant. The other reactants are called excess reactants. Also, the limiting reactant is completely consumed the initial mass of limiting reactant is used to calculate mass of product.

Blurred answer
Students have asked these similar questions
The Haber-Bosch process permits the direct conversion of molecular nitrogen to ammonia, which can be used in large-scale fertilizer production. Given the balanced Haber-Bosch reaction and using the bond energies in the table below, estimate the enthalpy change associated with the reaction. N2(g) + 3H2(g) → 2NH3(g) Bond N=N N = N Energy (kJ/mol) 941 418 N-N H-H N-H 163 435 388
Benzoic acid is used to determine the heat capacity of bomb calorimeters because it can be obtained in pure form and its energy of combustion is known very accurately (−26.43 kJ/g). Determine the heat capacity of a calorimeter that had a temperature increase of 9.199°C when 3.500 g of benzoic acid was used.
Given the standard enthalpies of formation for the following substances, determine the reaction enthalpy for the following reaction. 2N2H4(g) + 2NO2(g) → 3N2(g) + 4H2O(g) AHrxn ? kJ Substance AH in kJ/mol N2H4(g) +95.4 NO2(g) +33.1 H2O(g) -241.8

Chapter 11 Solutions

General Chemistry

Ch. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Prob. 11.13PCh. 11 - Prob. 11.14PCh. 11 - Prob. 11.15PCh. 11 - Prob. 11.16PCh. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Prob. 11.19PCh. 11 - Prob. 11.20PCh. 11 - Prob. 11.21PCh. 11 - Prob. 11.22PCh. 11 - Prob. 11.23PCh. 11 - Prob. 11.24PCh. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Prob. 11.27PCh. 11 - Prob. 11.28PCh. 11 - Prob. 11.29PCh. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - Prob. 11.32PCh. 11 - Prob. 11.33PCh. 11 - Prob. 11.34PCh. 11 - Prob. 11.35PCh. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Prob. 11.40PCh. 11 - Prob. 11.41PCh. 11 - Prob. 11.42PCh. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Prob. 11.47PCh. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Prob. 11.53PCh. 11 - Prob. 11.54PCh. 11 - Prob. 11.55PCh. 11 - Prob. 11.56PCh. 11 - Prob. 11.57PCh. 11 - Prob. 11.58PCh. 11 - Prob. 11.59PCh. 11 - Prob. 11.60PCh. 11 - Prob. 11.61PCh. 11 - Prob. 11.62PCh. 11 - Prob. 11.63PCh. 11 - Prob. 11.64PCh. 11 - Prob. 11.65PCh. 11 - Prob. 11.66PCh. 11 - Prob. 11.67PCh. 11 - Prob. 11.68PCh. 11 - Prob. 11.69PCh. 11 - Prob. 11.70PCh. 11 - Prob. 11.71PCh. 11 - Prob. 11.72PCh. 11 - Prob. 11.73PCh. 11 - Prob. 11.74PCh. 11 - Prob. 11.75PCh. 11 - Prob. 11.76PCh. 11 - Prob. 11.77PCh. 11 - Prob. 11.78PCh. 11 - Prob. 11.79PCh. 11 - Prob. 11.80PCh. 11 - Prob. 11.81PCh. 11 - Prob. 11.82PCh. 11 - Prob. 11.83PCh. 11 - Prob. 11.84PCh. 11 - Prob. 11.85PCh. 11 - Prob. 11.86PCh. 11 - Prob. 11.87PCh. 11 - Prob. 11.88PCh. 11 - Prob. 11.89PCh. 11 - Prob. 11.90PCh. 11 - Prob. 11.91PCh. 11 - Prob. 11.92PCh. 11 - Prob. 11.93P
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.
Similar questions
SEE MORE QUESTIONS
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