Potassium superoxide (KO₂, 71.10 g/mol) can be used to generate oxygen gas.If the theoretical yield of O₂ (32.00 g/mol) must be 12.0 g, how many grams of KO₂ should be used? Assume the reaction occurs in the presence of excess H₂O (18.02 g/mol) .
4 KO₂(s) + 2 H₂O(l) $\rightarrow$ 4 KOH(s) + 3O₂ (g)

Question

Potassium superoxide (KO₂, 71.10 g/mol) can be used to generate oxygen gas.If the theoretical yield of O₂ (32.00 g/mol) must be 12.0 g, how many grams of KO₂ should be used? Assume the reaction occurs in the presence of excess H₂O (18.02 g/mol) .
4 KO₂(s) + 2 H₂O(l)

\rightarrow

4 KOH(s) + 3O₂ (g)

Quizplus · Accepted Answer

The Answer of Potassium superoxide (KO₂, 71.10 g/mol) can be used to generate oxygen gas.If the theoretical yield of O₂ (32.00 g/mol) must be 12.0 g, how many grams of KO₂ should be used? Assume the reaction occurs in the presence of excess H₂O (18.02 g/mol). 4 KO₂(s) + 2 H₂O(l) $\rightarrow$ 4 KOH(s) + 3O₂ (g)

Potassium Superoxide (KO2, 71 →\rightarrow→ 4 KOH(s) + 3O2 (G) A)26

Potassium Superoxide (KO₂, 71 $\rightarrow$ 4 KOH(s) + 3O₂ (G)
A)26