Potassium superoxide (KO₂,71.10 g/mol)can be used to generate oxygen gas.What is the theoretical yield of O₂ (32.00 g/mol)if 500.0 g KO₂ reacts with excess H₂O (18.02 g/mol)? 4 KO₂(s)+ 2 H₂O $\to$ 4 KOH(s)+ 3 O₂(g)

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of KO₂ using its molar mass: 500.0 g KO₂ × (1 mol KO₂/71.10 g KO₂) = 7.035 mol KO₂ Since KO₂ is in excess, we can calculate the number of moles of O₂ using the stoichiometry of the reaction: 7.035 mol KO₂ × (3 mol O₂/4 mol KO₂) = 5.276 mol O₂ Finally, we can calculate the theoretical yield of O₂ using its molar mass: 5.276 mol O₂ × (32.00 g O₂/1 mol O₂) = 168.8 g O₂

Potassium Superoxide (KO2,71 →\to→ 4 KOH(s)+ 3 O2(g) A)225

Potassium Superoxide (KO₂,71 $\to$ 4 KOH(s)+ 3 O₂(g)
A)225