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 Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g. Experimental tests revealed that Na3PO4 • 12H2O (molar mass = 380.12 g/mol) was the limitin

Solution: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g. Experimental tests revealed that Na3PO4 • 12H2O (molar mass = 380.12 g/mol) was the limitin

Problem

The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g. Experimental tests revealed that Na3PO4 • 12H2O (molar mass = 380.12 g/mol) was the limiting reactant in the formation of the precipitate and the BaCl2 • 2H2O was the excess reactant in the salt mixture. Determine the mass of Na3PO4 • 12H2O in the salt mixture.

 

Solution
  • First step is to write the balanced equation, we need to find the other molecules produced when Na3PO4 • 12H2O is reacted with BaCl2 2H2O.

  • In the process NaCl and H2O is produced along with Ba3(PO4)2

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