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: Determine the theoretical yield of K2CO3 when 0.392 moles of KO2 react with 29.0 L of CO2 (at STP - 0°C, 1 atm). The molar mass of KO 2 = 71.10 g/mol and K2CO3 = 138.21 g/mol. 4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g) a) 27.1 gb) 179 gc) 91.7 gd) 206 ge) 61.0 g 

Problem

Determine the theoretical yield of K2CO3 when 0.392 moles of KO2 react with 29.0 L of CO2 (at STP - 0°C, 1 atm). The molar mass of KO 2 = 71.10 g/mol and K2CO3 = 138.21 g/mol.

 

4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

 

a) 27.1 g

b) 179 g

c) 91.7 g

d) 206 g

e) 61.0 g