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 $\to$ 4 KOH(s)+ 3 O₂(g)

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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 $\to$ 4 KOH(s)+ 3 O₂(g)

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

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