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: The zirconium oxalate K2Zr(C2O4)3(H2C2O4)·H2O was synthesized by mixing 1.68 g of ZrOCl2·8H2O with 5.20 g of H2C2O4·2H2O and an excess of aqueous KOH. After 2 months, 1.25 g of crystalline product was

Solution: The zirconium oxalate K2Zr(C2O4)3(H2C2O4)·H2O was synthesized by mixing 1.68 g of ZrOCl2·8H2O with 5.20 g of H2C2O4·2H2O and an excess of aqueous KOH. After 2 months, 1.25 g of crystalline product was

Problem

The zirconium oxalate K2Zr(C2O4)3(H2C2O4)·H2O was synthesized by mixing 1.68 g of ZrOCl2·8H2O with 5.20 g of H2C2O4·2H2O and an excess of aqueous KOH. After 2 months, 1.25 g of crystalline product was obtained, along with aqueous KCl and water. Calculate the percent yield.

Solution

The given reaction is:

We first need to balance this chemical equation. Add a coefficient of 2 to KCl to balance the number of Cl:

Add a coefficient of 4 to KOH to balance the number of K:

Add a coefficient of 4 to H2C2O4 to balance the number of C:

Add a coefficient of 20 to H2O to balance the number of H. This also balances the number of O. The balanced complete equation is:

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