Ch.4 - Chemical Quantities & Aqueous 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: Lead(Il) nitrate and ammonium iodide react to form lead(Il) iodide and ammonium nitrate according to the Pb(NO3)2 (aq) + 2NH4I (aq) → PbI2 (s) + 2NH4NO3 (aq) a) What volume of a 0.370 M NH 4l solution

Solution: Lead(Il) nitrate and ammonium iodide react to form lead(Il) iodide and ammonium nitrate according to the Pb(NO3)2 (aq) + 2NH4I (aq) → PbI2 (s) + 2NH4NO3 (aq) a) What volume of a 0.370 M NH 4l solution

Problem

Lead(Il) nitrate and ammonium iodide react to form lead(Il) iodide and ammonium nitrate according to the 

Pb(NO3)2 (aq) + 2NH4I (aq) → PbI2 (s) + 2NH4NO3 (aq) 

a) What volume of a 0.370 M NH 4l solution is required to react with 429 mL of a 0.140 M Pb(NO3)2 solution? 




b) How many moles of Pbl2 are formed from this reaction?