You would need 18 grams to get the requirement. Using a chemical equation can help to solve this for you.
18.8 G of KO2 are needed to form 8.5 grams of O2.
Approximately eighty-nine grams of KO2 are needed to prepare 100 grams of O2. This chemical equation is a prime example of how to determine the amount needed based on the molar mass.
19g
323.2 grams of KO2 is equal to 4.55 moles of KO2. Per the equation, 3.4125 moles of O2 is produced which is equal to 2.055 E24 molecules of oxygen. This means that every mg of KO2 produces 6.36 E18 molecules of O2.
Use.PV = nRTTo get moles O2. ( 20.0o C = 293.15 Kelvin )(1.00 atm)(1120.0 L) = (X moles)(0.08206 L*atm/mol*K)(293.15 K)Moles = 1120.0/24.56= 45.60 moles O2-----------------------------------now,45.60 moles O2 (1 mole KO2/2 mole O2)(71.1 grams/1 mole KO2)= 1621 grams potassium oxide required=============================
The thermal decomposition reaction is:2 KO2------------K2O2 + O20,2 moles of O2 are produced from o,4 moles KO2.
K + o2 = ko2
KO2
The principal oxidation state of NA in NA2O2 and K in KO2 is +1
+3
It is KO2 .
The reaction of carbon dioxide and potassium oxide is 4KO2 + 2CO2 = 2K2CO3 + 3O2. 156 grams of CO2 is 3.54 moles, which will produce 5.31 moles of O2.
KO2, [O2-] being the superoxide anion.
K2O: 3 atoms in the molecule K2O2: 4 atoms in the molecule KO2: 3 atoms in the molecule
Three potassium oxides are known: K2O, KO2, K2O2.