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: Sulfur and fluorine react to form sulfur hexafluoride:S(s) + 3 F2(g) → SF6(g)If 50.0 g S is allowed to react as completely as possible with 105.0 g F2(g), what mass of the excess reactant is left?

Solution: Sulfur and fluorine react to form sulfur hexafluoride:S(s) + 3 F2(g) → SF6(g)If 50.0 g S is allowed to react as completely as possible with 105.0 g F2(g), what mass of the excess reactant is left?

Problem

Sulfur and fluorine react to form sulfur hexafluoride:

S(s) + 3 F2(g) → SF6(g)

If 50.0 g S is allowed to react as completely as possible with 105.0 g F2(g), what mass of the excess reactant is left?

Solution

We first need to determine the moles of SF6 formed by each reactant. The reactant that forms the lower amount of product is the limiting reactant and will dictate the amount of product formed.

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