Ch.3 - Chemical ReactionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Assuming an efficiency of 30.60%, calculate the actual yield of magnesium nitrate formed from 140.2 g of magnesium and excess copper (II) nitrate.Mg + Cu(NO3)2 → Mg(NO3)2 + Cu

Solution: Assuming an efficiency of 30.60%, calculate the actual yield of magnesium nitrate formed from 140.2 g of magnesium and excess copper (II) nitrate.Mg + Cu(NO3)2 → Mg(NO3)2 + Cu

Problem

Assuming an efficiency of 30.60%, calculate the actual yield of magnesium nitrate formed from 140.2 g of magnesium and excess copper (II) nitrate.

Mg + Cu(NO3)2 → Mg(NO3)2 + Cu

Solution

We’re being asked to calculate the actual yield of Mg(NO3)2 in the reaction if the efficiency (percent yield) is 30.60%


The balanced chemical equation is:

Mg + Cu(NO3)2 Mg(NO3)2 + Cu


Recall that percent yield is given by:


%Yield=actual yieldtheoretical yield×100


The theoretical yield is not given but we can use the given information: 140.2 g Mg reacts with excess Cu(NO3)2.


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