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: Acrylonitrile (C3H3N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction:2 C3H6 (g) + 2 NH 3 (g) + 3 O2 (g) → 2 C3H3N (g) + 6 H2O (g)If 15.0 g C3

Solution: Acrylonitrile (C3H3N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction:2 C3H6 (g) + 2 NH 3 (g) + 3 O2 (g) → 2 C3H3N (g) + 6 H2O (g)If 15.0 g C3

Problem

Acrylonitrile (C3H3N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction:

2 C3H6 (g) + 2 NH 3 (g) + 3 O2 (g) → 2 C3H3N (g) + 6 H2O (g)

If 15.0 g C3H6, 10.0 g O 2, and 5.00 g NH 3 are reacted, what mass of acrylonitrile can be produced, assuming 100% yield?

Solution
  • Using the balanced equation provided, we can determine the mass of C3H3N based on the limiting reactant.
  • The limiting reactant is the reactant that will produce the least amount of C3H3N, we can determine this by using stoichiometry and the molar masses of each reactant:

MM of C3H6

C : 12.01 x 3 = 36.03

H: 1.01 x 6 = 6.06

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MM of C3H: 36.03 + 6.06  = 42.09 g/mol

MM of O2: 16 x 2 = 32 g/mol

MM of  NH3:

N: 14.01 x 1 = 14.01

H: 1.01 x 3 = 3.03 

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MM of NH3: 14.01 + 3.03 = 17.04 g/mol

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