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: Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al2S3 + H2O ⟶Al(OH)3 + H2S [unbalanced]How many grams of the excess rea

Solution: Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al2S3 + H2O ⟶Al(OH)3 + H2S [unbalanced]How many grams of the excess rea

Problem

Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al2S3 + H2O ⟶Al(OH)3 + H2S [unbalanced]

How many grams of the excess reactant remain?

Solution
  • Using stoichiometry and a balanced equation, we can find the moles and mass of H2S using the molar masses and mass of reactants given

  • The limiting reactant will tell how much moles of H2S will be produced 

  • Balancing the reaction, balance Al first (2 on both sides): 

Al2S+ H2O ⟶ 2Al(OH)3 + H2S

  • Balance S (3 on both sides)

Al2S+ H2O ⟶ 2Al(OH)3 + 3H2S

  • Balance H and O (6 O and 10 H atoms on both sides)

Al2S+ 6H2O ⟶ 2Al(OH)3 + 3H2S (balanced equation)

  • Calculating for the molar masses:

MM of H2S

H - 1.01 (2) = 2.02

S - 32.07 (1) = 32.07

2.02 + 32.07 = 34.09 g/mol

MM of Al2S3:

Al - 26.98 (2) = 53.96

S - 32.07 (3) = 96.21

53.96 + 96.21 = 150.17 g/mol

MM of H2O :

H - 1.01 (2) = 2.02

O - 16 (1) = 16

16 + 2.02 = 18.02 g/mol

  • Calculating for the moles of H2S produced by the reactants:

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